Real Analog Chapter 10: Steady-state Sinusoidal Analysis ...

1300 Henley Court Pullman, WA 99163

509.334.6306 www.store. digilent.com

Real Analog Chapter 10: Steady-state Sinusoidal Analysis

Chapter 10 Copyright Digilent, Inc. All rights reserved. Other product and company names mentioned may be trademarks of their respective owners. Page 1 of 85

10 Introduction and Chapter Objectives

We will now study dynamic systems which are subjected to sinusoidal forcing functions. Previously, in our analysis

of dynamic systems, we determined both the unforced response (or homogeneous solution) and the forced

response (or particular solution) to the given forcing function. In the next several chapters, however, we will

restrict our attention to only the system’s forced response to a sinusoidal input; this response is commonly called

the sinusoidal steady-state system response. This analysis approach is useful if we are concerned primarily with

the system’s response after any initial conditions have died out, since we are ignoring any transient effects due to

the system’s natural response.

Restricting our attention to the steady-state sinusoidal response allows a considerable simplification in the system

analysis: we can solve algebraic equations rather than differential equations. This advantage often more than

compensates for the loss of information relative to the systems natural response. For example it is often the case

that a sinusoidal input is applied for a very long time relative to the time required for the natural response to die

out, so that the overall effects of the initial conditions are negligible.

Steady-state sinusoidal analysis methods are important for several reasons:

• Sinusoidal inputs are an extremely important category of forcing functions. In electrical engineering, for

example, sinusoids are the dominant signal in the electrical power industry. The alternating current (or

AC) signals used in power transmission are, in fact, so pervasive that many electrical engineers commonly

refer to any sinusoidal signal as “AC”. Carrier signals used in communications systems are also sinusoidal

in nature.

• The simplification associated with the analysis of steady state sinusoidal analysis is often so desirable that

system responses to non-sinusoidal inputs are interpreted in terms of their sinusoidal steady-state

response. This approach will be developed when we study Fourier series.

• System design requirements are often specified in terms of the desired steady-state sinusoidal response

of the system.

In section 10.1 of this chapter, we qualitatively introduce the basic concepts relative to sinusoidal steady state

analyses so that readers can get the “general idea” behind the analysis approach before addressing the

mathematical details in later sections. Since we will be dealing exclusively with sinusoidal signals for the next few

chapters, section 10.2 provides review material relative to sinusoidal signals and complex exponentials. Recall

from chapter 8 that complex exponentials are a mathematically convenient way to represent sinusoidal signals.

Most of the material in section 10.2 should be review, but the reader is strongly encouraged to study section 10.2

carefully -- we will be using sinusoids and complex exponentials extensively throughout the remainder of this text,

and a complete understanding of the concepts and terminology is crucial. In section 10.3, we examine the forced

response of electrical circuits to sinusoidal inputs; in this section, we analyze our circuits using differential

equations and come to the important conclusion that steady-state response of a circuit to sinusoidal inputs is


Copyright Digilent, Inc. All rights reserved. Other product and company names mentioned may be trademarks of their respective owners. Page 2 of 85

governed by algebraic equations. Section 10.4 takes advantage of this conclusion to perform steady-state

sinusoidal analyses of electrical circuits without writing the governing differential equation for the circuit! Finally,

in section 10.5, we characterize a system’s response purely by its effect on a sinusoidal input. This concept will be

used extensively throughout the remainder of this textbook.

After completing this chapter, you should be able to:

State the relationship between the sinusoidal steady state system response and the forced response of a

system

For sinusoidal steady-state conditions, state the relationship between the frequencies of the input and

output signals for a linear, time-invariant system

State the two parameters used to characterize the sinusoidal steady-state response of a linear, time-

invariant system

Define periodic signals

Define the amplitude, frequency, radian frequency, and phase of a sinusoidal signal

Express sinusoidal signals in phasor form

Perform frequency-domain analyses of electrical circuits

Sketch phasor diagrams of a circuit’s input and output

State the definition of impedance and admittance

State, from memory, the impedance relations for resistors, capacitors, and inductors

Calculate impedances for resistors, capacitors, and inductors

State how to use the following analysis approaches in the frequency domain:

o KVL and KCL

o Voltage and current dividers

o Circuit reduction techniques

o Nodal and mesh analysis

o Superposition, especially when multiple frequencies are present

o Thévenin’s and Norton’s theorems

Determine the load impedance necessary to deliver maximum power to a load

Define the frequency response of a system

Define the magnitude response and phase response of a system

Determine the magnitude and phase responses of a circuit

10.1 Introduction to Steady-state Sinusoidal Analysis

In this chapter, we will be almost exclusively concerned with sinusoidal signals, which can be written in the form:

𝑓(𝑡) = 𝐴 𝑐𝑜𝑠(𝜔𝑡 + 𝜃) Eq. 10.1

Where A is the amplitude of the sinusoid, ω is the angular frequency (in radians/second) of the signal, and θ is

the phase angle (expressed in radians or degrees) of the signal. A provides the peak value of the

sinusoid, ω governs the rate of oscillation of the signal, and θ affects the translation of the sinusoid in time. A

typical sinusoidal signal is shown in Fig. 10.1.



A

f(t)

Time, t

2

Figure 10.1. Sinusoidal signal.

If the sinusoidal signal of Fig. 10.1 is applied to a linear time invariant system, the response of the system will

consist of the system’s natural response (due to the initial conditions on the system) superimposed on the

system’s forced response (the response due to the forcing function). As we have seen in previous chapters, the

forced response has the same form as the forcing function. Thus, if the input is a constant value the forced

response is constant, as we have seen in the case of the step response of a system. In the case of a sinusoidal input

to a system, the forced response will consist of a sinusoid of the same frequency as the input sinusoid. Since the

natural response of the system decays with time, the steady state response of a linear time invariant system to a

sinusoidal input is a sinusoid, as shown in Fig. 10.2. The amplitude and phase of the output may be different than

the input amplitude and phase, but both the input and output signals have the same frequency.

It is common to characterize a system by the ratio of the magnitudes of the input and output signals (𝐵

𝐴 in Fig. 10.2)

and the difference in phases between the input and output signals (ϕ−θ) in Fig. 10.2) at a particular frequency. It is

important to note that the ratio of magnitudes and difference in phases is dependent upon the frequency of the

applied sinusoidal signal.

SystemInput

u(t)=Acos(t+)

Output

y(t)=Bcos(t+f)

Figure 10.2. Sinusoidal steady-state input-output relation for a linear time invariant system.

Example 10.1: Series RLC Circuit Response

Consider the series RLC circuit shown in Fig. 10.3 below. The input voltage to the circuit is given by:

𝑣𝑠(𝑡) = 0, 𝑡 < 0

cos(5𝑡) , 𝑡 ≥ 0

Thus, the input is zero prior to t=0, and the sinusoidal input is suddenly “switched on” at time t=0. The input

forcing function is shown in Fig. 10.4(a). The circuit is “relaxed” before the sinusoidal input is applied, so the circuit

initial conditions are:

𝑦(0−) =𝑑𝑦

𝑑𝑡|𝑡=0− = 0



+

-vs(t)

0.004 F1 W

y(t)+ -

1 H

Figure 10.3. Series RLC circuit; output is voltage across capacitor.

This circuit has been analyzed previously in Chapter 8, and the derivation of the governing differential equation will

not be repeated here. The full output response of the circuit is shown in Fig. 10.4(b). The natural response of the

circuit is readily apparent in the initial portion of the response but these transients die out quickly, leaving only the

sinusoidal steady-state response of the circuit. It is only this steady state response in which we will be interested

for the next several modules. With knowledge of the frequency of the signals, we can define both the input and

(steady-state) output by their amplitude and phase, and characterize the circuit by the ratio of the output-to-input

amplitude and the difference in the phases of the output and input.

time

u(t)

(a) Input signal

time

y(t)

Steady-State Response

(b) Output signal.

Figure 10.4. Input and output signals for circuit of Figure 10.3.

Section Summary

Sinusoidal signals can be expressed mathematically in the form:



𝑓(𝑡) = 𝐴 cos (𝜔𝑡 + 𝜃)

• In the above, A is the amplitude of the sinusoid, it describes the maximum and minimum values of the

signal.

• In the above, θ is the phase angle of the sinusoid, it describes the time shift of the sinusoid relative to a

pure cosine.

• In the above, ω is the radian frequency of the sinusoid. The sinusoid repeats itself at time intervals

of 2𝜋

𝜔 seconds.

• A sinusoidal signal is completely described by its frequency, its amplitude, and its phase angle.

• The steady-state response of a linear, time-invariant system to a sinusoidal input is a sinusoid with the

same frequency.

• Since the frequencies of the input and output are the same, the relationship between the input and

output sinusoids is completely characterized by the relationships between:

o The input and output amplitudes.

o The input and output phase angles.

10.1 Exercises

1. In the circuit below, all circuit elements are linear and time invariant. The input voltage 𝑉𝑖𝑛(𝑡) =

10 cos (2𝑡 + 40°). What is the radian frequency of the output voltage 𝑉𝑜𝑢𝑡(𝑡)?

Vin(t)+-

+

-

Vout(t)

2. In the circuit below, all circuit elements are linear and time invariant. The input voltage is𝑉𝑖𝑛(𝑡) =

10 cos (2𝑡 + 40°). The output voltage is of the form 𝑉𝑜𝑢𝑡(𝑡) = 𝐴 cos (𝜔𝑡 + 𝜙°). If the ratio between the

input and output, |𝑉𝑜𝑢𝑡

𝑉𝑖𝑛| = 0.5 and the phase difference between the input and output is 20, what are:

a. The radian frequency of the output, ?

b. The amplitude of the output, A?

c. The phase angle of the output, f?

Vin(t)+-

+ -Vout(t)



10.2 Sinusoidal Signals, Complex Exponentials, and Phasors

In this section, we will review properties of sinusoidal functions and complex exponentials. We will also introduce

phasor notation, which will significantly simplify the sinusoidal steady-state analysis of systems, and provide

terminology which will be used in subsequent sinusoidal steady-state related modules.

Much of the material presented here has been provided previously in Chapter 8; this material is, however,

important enough to bear repetition. Likewise, a brief overview of complex arithmetic, which will be essential in

using complex exponentials effectively, is provided at the end of this section. Readers who need to review

complex arithmetic may find it useful to peruse this overview before reading the material in this section relating to

complex exponentials and phasors.

10.2.1 Sinusoidal Signals

The sinusoidal signal shown in Fig. 10.5 is represented mathematically by:

𝑓(𝑡) = 𝑉𝑃 cos(𝜔𝑡) Eq. 10.2

The amplitude or peak value of the function is VP. VP is the maximum value achieved by the function; the function

itself is bounded by +VP and −VP, so that -VP≤f(t)≤VP. The radian frequency or angular frequency of the function

is ω; the units of ω are radians/second. The function is said to be periodic; periodic functions repeat themselves at

regular intervals, so that:

𝑓(𝑡 + 𝑛𝑇) = 𝑓(𝑡) Eq. 10.3

Where n is any integer and T is the period of the signal. The sinusoidal waveform shown in Fig. 10.5 goes through

one complete cycle or period in T seconds. Since the sinusoid of equation (10.2) repeats itself every 2π radians,

the period is related to the radian frequency of the sinusoid by:

𝜔 =2𝜋

𝑇 Eq. 10.4

It is common to define the frequency of the sinusoid in terms of the number of cycles of the waveform which occur

in one second. In these terms, the frequency f of the function is:

𝑓 =1

𝑇 Eq. 10.5

The units of f are cycles/second or Hertz (abbreviated Hz). The frequency and radian frequency are related by:

𝑓 =𝜔

2𝜋 Eq. 10.6

Or equivalently:

𝜔 = 2𝜋𝑓 Eq. 10.7

Regardless of whether the sinusoid’s rate of oscillation is expressed as frequency or radian frequency, it is

important to realize that the argument of the sinusoid in equation (10.2) must be expressed in radians. Thus,

equation (10.2) can be expressed in terms of frequency in Hz as:

𝑓(𝑡) = cos(2𝜋𝑓𝑡) Eq. 10.8



To avoid confusion in our mathematics, we will almost invariably write sinusoidal functions in terms of radian

frequency as shown in equation (10.2), although Hz is generally taken as the standard unit for frequency

(experimental apparatus, for example, commonly express frequency in Hz).

VP

-VP

T

f(t)

t, sec

Figure 10.5. Pure cosine waveform.

A more general expression of a sinusoidal signal is:

𝑣(𝑡) = 𝑉𝑃cos (𝜔𝑡 + 𝜃) Eq. 10.9

Where θ is the phase angle or phase of the sinusoid. The phase angle simply translates the sinusoid along the time

axis, as shown in Fig. 10.6. A positive phase angle shifts the signal left in time, while a negative phase angle shifts

the signal right – this is consistent with our discussion of step functions in section 6.1, where it was noted that

subtracting a value from the unit step argument resulted a time delay of the function. Thus, as shown in Figure

10.6, a positive phase angle causes the sinusoid to be shifted left by θω seconds.

The units of phase angle should be radians, to be consistent with the units of ωt in the argument of the cosine. It is

typical, however, to express phase angle in degrees, with 180∘ corresponding to π radians. Thus, the conversion

between radians and degrees can be expressed as:

Number of degrees =180

𝜋𝑥 Number of radians

For example, we will consider the two expressions below to be equivalent, though the expression on the right-

hand side of the equal sign contains a mathematical inconsistency:

𝑉𝑃 cos (𝜔𝑡 +𝜋

2) = 𝑉𝑃cos (𝜔𝑡 + 90°)



VP

-VPT

t, sec

v(t)

Figure 10.6. Cosine waveform with non-zero phase angle.

For convenience, we introduce the terms leading and lagging when referring to the sign on the phase angle, θ. A

sinusoidal signal v1(t) is said to lead another sinusoid v2(t) of the same frequency if the phase difference between

the two is such that v1(t) is shifted left in time relative to v2(t). Likewise, v1(t) is said to lag another sinusoid v2(t) of

the same frequency if the phase difference between the two is such that v1(t) is shifted right in time relative

to v2(t). This terminology is described graphically in Fig. 10.7.

cos(t+)

< 0

lags cos(t)

Time

v(t)cos(t+)

> 0

leads cos(t)

cos(t)

Figure 10.7. Leading and lagging sinusoids.

Finally, we note that the representation of sinusoidal signals as a phase shifted cosine function, as provided by

equation (10.9), is completely general. If we are given a sinusoidal function in terms of a sine function, it can be

readily converted to the form of equation (10.9) by subtracting a phase of 𝜋

2 (or 90 ) from the argument, since:

sin(𝜔𝑡) = cos (𝜔𝑡 −𝜋

2)

Likewise, sign changes can be accounted for by a ±π radian phase shift, since:

− cos(𝜔𝑡) = cos(𝜔𝑡 ± 𝜋)

Obviously, we could have chosen either a cosine or sine representation of a sinusoidal signal. We prefer the cosine

representation, since a cosine is the real part of a complex exponential. In the next module, we will see that

sinusoidal steady-state circuit analysis is simplified significantly by using complex exponentials to represent the

sinusoidal functions. The cosine is the real part of a complex exponential (as we saw previously in chapter 8). Since

all measurable signals are real valued, we take the real part of our complex exponential-based result as our

physical response; this results in a solution of the form of equation (10.9).



Since representation of sinusoidal waveforms as complex exponentials will become important to us in circuit

analysis, we devote the following subsection to a review of complex exponentials and their interpretation as

sinusoidal signals.

10.2.2 Complex Exponentials and Phasors

Euler’s identity can be used to represent complex numbers as complex exponentials:

𝑒𝑗𝜃 = cos 𝜃 ± 𝑗 sin 𝜃 Eq. 10.10

If we generalize equation (9) to time-varying signals of arbitrary magnitude, we can write:

𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃) = 𝑉𝑃 cos(𝜔𝑡 + 𝜃) ± 𝑗𝑉𝑃 sin(𝜔𝑡 + 𝜃) Eq. 10.11

So that:

𝑉𝑃 cos(𝜔𝑡 + 𝜃) = 𝑅𝑒𝑉𝑃𝑒±(𝜔𝑡+𝜃) Eq. 10.12

And:

𝑉𝑃 sin(𝜔𝑡 + 𝜃) = 𝐼𝑚𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃) Eq. 10.13

Where 𝑅𝑒𝑉𝑃𝑒±(𝜔𝑡+𝜃) and 𝐼𝑚𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃) denote the real part of 𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃) and the imaginary part of

𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃), respectively. The complex exponential of equation (10.11) can also be written as:

𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃) = 𝑉𝑃𝑒𝑗𝜃𝑒𝑗𝜔𝑡 Eq. 10.14

The term 𝑉𝑃𝑒𝑗𝜃 on the right-hand side of equation (10.14) is simply a complex number which provides the

magnitude and phase information of the complex exponential of equation (10.11). From equation (10.12), this

magnitude and phase can be used to express the magnitude and phase angle of a sinusoidal signal of the form

given in equation (10.9).

The complex number in polar coordinates which provides the magnitude and phase angle of a time-varying

complex exponential, as given in equation (10.14) is called a phasor. The phasor representing 𝑉𝑃cos (𝜔𝑡 + 𝜃) is

defined as:

𝑉 = 𝑉𝑃𝑒𝑗𝜃 = 𝑉𝑃∠𝜃 Eq. 10.15

We will use a capital letter with an underscore to denote a phasor. Using bold typeface to represent phasors is

more common; our notation is simply for consistency between lecture material and written material – boldface

type is difficult to create on a whiteboard during lecture!

Note: The phasor representing a sinusoid does not provide information about the frequency of the sinusoid

– frequency information must be kept track of separately.

10.2.3 Complex Arithmetic Review

Much the material in this section has been provided previously in section 8.3. It is repeated here to emphasize its

importance and to expand slightly upon some crucial topics.

In our presentation of complex exponentials, we first provide a brief review of complex numbers. A complex

number contains both real and imaginary parts. Thus, we may write a complex number 𝐴 as:

𝐴𝑎 + 𝑗𝑏 Eq. 10.16



Where:

𝑗 = √−1 Eq. 10.17

And the underscore denotes a complex number. The complex number 𝐴 can be represented on orthogonal axes

representing the real and imaginary part of the number, as shown in Fig. 10.8. (In Figure 10.8, we have taken the

liberty of representing 𝐴 as a vector, although it is really just a number.) We can also represent the complex

number in polar coordinates, also shown in Figure 10.8. The polar coordinates consist of a magnitude |A| and

phase angle θA, defined as:

|𝐴| = √𝑎2 + 𝑏2 Eq. 10.18

𝜃𝐴 = tan−1 (𝑏

𝑎) Eq. 10.19

Notice that the phase angle is defined counterclockwise from the positive real axis. Conversely, we can determine

the rectangular coordinates from the polar coordinates from:

𝑎 = 𝑅𝑒𝐴 = |𝐴| cos(𝜃𝐴) Eq. 10.20

𝑏 = 𝐼𝑚𝐴 = |𝐴|sin (𝜃𝐴) Eq. 10.21

Where the notation 𝑅𝑒𝐴 and 𝐼𝑚𝐴 denote the real part of 𝐴 and the imaginary part of 𝐴, respectively.

The polar coordinates of a complex number of 𝐴 are often represented in the form:

𝐴|𝐴|∠𝜃𝐴 Eq. 10.22

A

A

Re

Im

a

b

)cos(A A

)sin(A A

Figure 10.8. Representation of a complex number in rectangular and polar coordinates.

An alternate method of representing complex numbers in polar coordinates employs complex exponential

notation. Without proof, we claim that:

𝑒𝑗𝜃 = 1∠𝜃 Eq. 10.23

Thus, 𝑒𝑗𝜃 is a complex number with magnitude 1 and phase angle θ. From Fig. 10.8, it is easy to see that this

definition of the complex exponential agrees with Euler’s equation:

𝑒±𝑗𝜃 = 𝑐𝑜𝑠𝜃 ± 𝑗 sin 𝜃 Eq. 10.24

With the definition of equation (10.23), we can define any arbitrary complex number in terms of complex

numbers. For example, our previous complex number 𝐴 can be represented as:

𝐴 = |𝐴|𝑒𝑗𝜃𝐴 Eq. 10.25



We can generalize our definition of the complex exponential to time-varying signals. If we define a time varying

signal 𝑒𝑗𝜔𝑡, we can use equation (10.24) to write:

𝑒𝑗𝜔𝑡 = cos 𝜔𝑡 ± 𝑗 sin 𝜔𝑡 Eq. 10.26

The signal 𝑒𝑗𝜔𝑡 can be visualized as a unit vector rotating around the origin in the complex plane; the tip of the

vector scribes a unit circle with its center at the origin of the complex plane. This is illustrated in Fig. 10.9. The

vector rotates at a rate defined by the quantity ω– the vector makes one complete revolution every 2𝜋

𝜔 seconds.

The projection of this rotating vector on the real axis traces out the signal cos 𝜔𝑡, as shown in Fig. 10.7, while the

projection of the rotating vector on the imaginary axis traces out the signal sin 𝜔𝑡, also shown in Fig. 10.9.

Thus, we interpret the complex exponential function 𝑒𝑗𝜔𝑡 as an alternate “type” of sinusoidal signal. The real part

of this function is cos 𝜔𝑡 while the imaginary part of this function is sin 𝜔𝑡.

Im

Re

t

cos t

sin t

time

time

t

t

Figure 10.9. Illustration of tje

.

Addition and subtraction of complex numbers is most easily performed in rectangular coordinates. Given two

complex numbers 𝐴 and 𝐵, defined as:

𝐴 = 𝑎 + 𝑗𝑏

𝐵 = 𝑐 + 𝑗𝑑

The sum and difference of the complex number can be determined by:

𝐴 + 𝐵 = (𝑎 + 𝑐) + 𝑗(𝑏 + 𝑑)

And:

𝐴 − 𝐵 = (𝑎 − 𝑐) + 𝑗(𝑏 − 𝑑)



Multiplication and division, on the other hand, are probably most easily performed using polar coordinates. If we

define two complex numbers as:

𝐴 = |𝐴|𝑒𝑗𝜃𝐴 = |𝐴|∠𝜃𝐴

𝐵 = |𝐵|𝑒𝑗𝜃𝐵 = |𝐵|∠𝜃𝐵

The product and quotient can be determined by:

𝐴 ⋅ 𝐵 = |𝐴|𝑒𝑗𝜃𝐴 ⋅ |𝐵|𝑒𝑗𝜃𝐵 = |𝐴| ⋅ |𝐵|𝑒𝑗(𝜃𝐴+𝜃𝐵) = |𝐴| ⋅ |𝐵|∠(𝜃𝐴 + 𝜃𝐵)

And:

𝐴

𝐵=

|𝐴|𝑒𝑗𝜃𝐴

|𝐵|𝑒𝑗𝜃𝐵=

𝐴

𝐵∠(𝜃𝐴 − 𝜃𝐵)

The conjugate of a complex number, denoted by a *, is obtained by changing the sign on the imaginary part of the

number. For example, if 𝐴 = 𝑎 + 𝑗𝑏 = |𝐴|𝑒𝑗𝜃, then:

𝐴∗ = 𝑎 − 𝑗𝑏 = |𝐴|𝑒−𝑗𝜃

Conjugation does not affect the magnitude of the complex number, but it changes the sign on the phase angle. It is

easy to show that:

𝐴 ⋅ 𝐴∗ = |𝐴|2

Several useful relationships between polar and rectangular coordinate representations of complex numbers are

provided below. The reader is encouraged to prove any that are not self-evident.

𝑗 = 1∠90°

−𝑗 = 1∠ − 90° 1

𝑗= −𝑗 = 1∠ − 90°

1 = 1∠0°

−1 = 1∠180°

Section Summary

• Periodic signals repeat themselves at a specific time interval. Sinusoidal signals are a special case of

periodic signals. • A sinusoidal signal can always be written in the form v(t) = VP cos(ωt + θ).

• It is often convenient, when analyzing a system’s steady-state response to sinusoidal inputs, to express

sinusoidal signals in terms of complex exponentials. This is possible because of Euler’s formula:

𝑒±𝑗𝜔𝑡 = cos 𝜔𝑡 ± 𝑗 sin 𝜔𝑡

• From Euler’s formula, a sinusoidal signal can be expressed as the real part of a complex exponential:

𝑣(𝑡) = 𝑉𝑃 cos(𝜔𝑡 + 𝜃) = 𝑅𝑒𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃)

• The magnitude and phase angle of a complex exponential signal are conveniently expressed as a phasor:

𝑉 = 𝑉𝑃𝑒𝑗𝜃



• Using phasor notation, the above complex exponential signal can be written as:

𝑉𝑃𝑒±𝑗(𝜔𝑡+𝜃) = 𝑉𝑒𝑗𝜔𝑡

• Phasors can then be operated on arithmetically in the same way as any other complex number. However,

when operating on phasors, keep in mind that you are dealing with the amplitude and phase angle of a

sinusoidal signal.

10.2 Exercises

1. Express the following complex numbers in rectangular form:

1.1. 3𝑒𝑗45°

1.2. 5√2𝑒𝑗135°

1.3. 2.5𝑒𝑗90°

1.4. 6𝑒𝑗𝜋

2. Express the following complex numbers in complex exponential form:

2.1. 2 − 𝑗2

2.2. −𝑗3

2.3. 6

2.4. 3 + 𝑗

3. Evaluate the following expressions. Express your results in complex exponential form.

3.1. −𝑗

2(𝑗+1)

3.2. 2−𝑗2

4+𝑗4

3.3. 2𝑒𝑗45° ⋅2

𝑗+1

3.4. 𝑗 +2

𝑗

4. Represent the following sinusoids in phasor form:

4.1. 3 cos(5𝑡 − 60°)

4.2. −2 cos(300𝑡 + 45°)

4.3. sin (6𝑡)

4.4. 7 cos(3𝑡)

5. Write the signal representing the real part of the following complex exponentials:

5.1. 5√2𝑒𝑗(100𝑡−45°)

5.2. 3𝑒𝑗𝜋𝑒𝑗3𝑡

5.3. 2𝑒𝑗(𝜋𝑡−30°) + 4𝑒𝑗(4𝑡+20°)

10.3 Sinusoidal Steady-state System Response

In this section, the concepts presented in sections 10.1 and 10.2 are used to determine the sinusoidal steady-state

response of electrical circuits. We will develop sinusoidal steady-state circuit analysis in terms of examples, rather

than attempting to develop a generalized approach à priori. The approach is straightforward, so that a general

analysis approach can be inferred from the application of the method to several simple circuits.

The overall approach to introducing sinusoidal steady-state analysis techniques used in this section is as follow:



We first determine the sinusoidal steady-state response of a simple RC circuit, by solving the differential

equation governing the system. This results directly in a solution which is a function of time; it is a time

domain analysis technique. The approach is mathematically tedious, even for the simple circuit being

analyzed.

We then re-analyze the same RC circuit using complex exponentials and phasors. This approach results in

the transformation of the governing time domain differential equation into an algebraic equation which is

a function of frequency. It is said to describe the circuit behavior in the frequency domain. The frequency

domain equation governing the system is then solved using phasor techniques and the result transformed

back to the time domain. This approach tends to be mathematically simpler than the direct solution of the

differential equation in the time domain, though in later sections we will simplify the approach even

further.

Several other examples of sinusoidal steady-state circuit analysis are then performed using frequency

domain techniques in order to demonstrate application of the approach to more complex circuits. It will

be seen that, unlike time-domain analysis, the difficulty of the frequency domain analysis does not

increase drastically as the circuit being analyzed becomes more complex.

Example 10.2: RC Circuit Sinusoidal Steady-state Response via Time-domain

Analysis

In the circuit below, the input voltage is 𝑢(𝑡) = 𝑉𝑃 cos(𝜔𝑡) volts and the circuit response (or output) is the

capacitor voltage, y(t). We want to find the steady-state response (as 𝑡 → ∞).

+

-u(t) = Vpcos(t) C

R

y(t)

+

-

The differential equation governing the circuit is:

𝑑𝑦(𝑡)

𝑑𝑡+

1

𝑅𝐶𝑦(𝑡) =

𝑉+𝑃

𝑅𝐶 cos(𝜔𝑡) Eq. 10.27

Since we are concerned only with the steady-state response, there is no need to determine the homogeneous

solution of the differential equation (or, equivalently, the natural response of the system) so we will not be

concerned with the initial conditions on the system – their effect will have died out by the time we are interested

in the response. Thus, we only need to determine the particular solution of the above differential equation (the

forced response of the system). Since the input function is a sinusoid, the forced response must be sinusoidal, so

we assume that the forced response yf(t) has the form:

𝑦𝑓(𝑡) = 𝐴 cos(𝜔𝑡) + 𝐵 sin(𝜔𝑡) Eq. 10.28

Substituting equation (10.28) into equation (10.27) results in:

−𝐴𝜔 sin(𝜔𝑡) + 𝐵𝜔 cos(𝜔𝑡) +1

𝑅𝐶[cos(𝜔𝑡) + 𝐵𝑠𝑖𝑛 (𝜔𝑡)] =

𝑉𝑃

𝑅𝐶cos(𝜔𝑡) Eq. 10.29

Equating coefficients on the sine and cosine terms results in two equations in two unknowns:

−𝐴𝜔 +𝐵

𝑅𝐶= 0 𝐵𝜔 +

𝐴

𝑅𝐶=

𝑉𝑃

𝑅𝐶 Eq. 10.30



Solving equations (10.30) results in:

𝐴 =𝑉𝑃

1+(𝜔𝑅𝐶)2 𝐵 =𝑉𝑃𝜔𝑅𝐶

1+(𝜔𝑅𝐶)2 Eq. 10.31

Substituting equations (10.31) into equation (10.28) and using the trigonometric identity 𝐴 cos(𝜔𝑡) +

𝐵 sin(𝜔𝑡) = √𝐴2 + 𝐵2 cos [𝜔𝑡 − tan−1 (𝐵

𝐴)] results in (after some fairly tedious algebra):

𝑦𝑓(𝑡) =𝑉𝑃

√1+(𝜔𝑅𝐶)2cos[𝜔𝑡 − tan−1(𝜔𝑅𝐶)] Eq. 10.32

Note: In all steps of the above analysis, the functions being used are functions of time. That is, for a particular

value of ω, the functions vary with time. The above analysis is being performed in the time domain.

Example 10.3: RC Circuit Sinusoidal Steady-state Response via Frequency-domain

Analysis

We now repeat Example 10.2, using phasor-based analysis techniques. The circuit being analyzed is shown in the

figure to the left below for reference; the input voltage is 𝑢(𝑡) = 𝑉𝑃 cos(𝜔𝑡) volts and the circuit response (or

output) is the capacitor voltage, y(t). We still want to find the steady-state response (as 𝑡 → ∞). In this example,

we replace the physical input, 𝑢(𝑡) = 𝑉𝑃 cos(𝜔𝑡), with a conceptual input based on a complex exponential as

shown in the figure to the right below. The complex exponential input is chosen such that the real part of the

complex input is equivalent to the physical input applied to the circuit. We will analyze the conceptual circuit with

the complex valued input.

+

-u(t) = Vpcos(t) C

R

y(t)

+

-

+

-u(t) = Vpe

jt C

R

y(t)

+

-

The differential equation governing the circuit above is the same as in example 10.2, but with the complex input:

𝑑𝑦(𝑡)

𝑑𝑡+

1

𝑅𝐶𝑦(𝑡) =

𝑉𝑃

𝑅𝐶𝑒𝑗𝜔𝑡 Eq. 10.33

As in example 10.2, we now assume a form of the forced response. In this case, however, our solution will be

assumed to be a complex exponential:

𝑦(𝑡) = |𝑌|𝑒𝑗(𝜔𝑡+𝜃) Eq. 10.34

Which can be written in phasor form as:

𝑦(𝑡) = 𝑌𝑒𝑗𝜔𝑡 Eq. 10.35

Where the phasor 𝑌 is a complex number which can be expressed in either exponential or polar form:

𝑦(𝑡) = 𝑌𝑒𝑗𝜔𝑡 Eq. 10.36

Substituting (10.35) into equation (10.33) and taking the appropriate derivative results in:



𝑗𝜔𝑌𝑒𝑗𝜔𝑡 +1

𝑅𝐶𝑌𝑒𝑗𝜔𝑡 =

𝑉𝑃

𝑅𝐶𝑒𝑗𝜔𝑡 Eq. 10.37

We can divide equation (10.37) by 𝑒𝑗𝜔𝑡 to obtain:

𝑗𝜔𝑌 +1

𝑅𝐶𝑌 =

𝑉𝑃

𝑅𝐶 Eq. 10.38

Equation (10.38) can be solved for 𝑌:

(𝑗𝜔 +1

𝑅𝐶) 𝑌 =

𝑉𝑃

𝑅𝐶⇒ 𝑌 =

𝑉𝑃𝑅𝐶

𝑗𝜔+1

𝑅𝐶

Eq. 10.39

So that:

𝑌 =𝑉𝑃

1+𝑗𝜔𝑅𝐶 Eq. 10.40

The magnitude and phase of the output response can be determined from the phasor 𝑌:

|𝑌| =𝑉𝑃

√1+(𝜔𝑅𝐶)2

∠𝑌 = − tan−1(𝜔𝑅𝐶) Eq. 10.41

The complex exponential form of the system response is then, from equation (10.35):

𝑦(𝑡) =𝑉𝑃

√1+(𝜔𝑅𝐶)2𝑒𝑗(𝜔𝑡−tan−1(𝜔𝑅𝐶)) Eq. 10.42

Since our physical input is the real part of the conceptual input, and since all circuit parameters are real valued, our

physical output is the real part of equation (10.42) and the forced response is:

𝑦𝑓(𝑡) =𝑉𝑃

√1+(𝜔𝑅𝐶)2cos[𝜔𝑡 − tan−1(𝜔𝑅𝐶)] Eq. 10.43

Which agrees with our result from the time-domain analysis of example 10.2.

Notes:

The transition from equation (10.37) to equation (10.38) removed the time-dependence of our solution.

The solution is now no longer a function of time! The solution includes the phasor representations of the

input and output, as well as (generally) frequency. Thus, equation (10.38) is said to be in the phasor

domain or, somewhat more commonly, the frequency domain. The analysis remains in the frequency

domain until we reintroduce time in equation (10.43).

Equations in the frequency domain are algebraic equations rather than differential equations. This is a

significant advantage mathematically, especially for higher-order systems.

Circuit components must have purely real values for the above process to work. We do not prove this, but

merely make the claim that the process of taking the real part of the complex exponential form of the

system response is not valid if circuit components (or any coefficients in the differential equation

governing the system) are complex valued. Fortunately, this is not a strong restriction – complex values do

not exist in the physical world.

The complex exponential we use for our “conceptual” input, 𝑉𝑃𝑒𝑗𝜔𝑡 , is not physically realizable. That is,

we cannot create this signal in the real world. It is a purely mathematical entity which we introduce solely

for the purpose of simplifying the analysis. The complex form of the output response given by equation

(10.42) is likewise not physically realizable.



Example 10.4: Numerical Example and Phasor Diagrams

We now examine the circuit shown below. This circuit is simply the circuit of Example 10.3, with 𝑅 = 1𝑘Ω, 𝐶 =

1𝜇𝐹, 𝑉𝑃 = 5𝑉, and 𝜔 = 1000 rad/second.

+

-u(t) = 5cos(1000t) 1 mF

1 kW

y(t)

+

-

In phasor form, the input is 𝑢(𝑡) = 𝑈𝑒𝑗1000𝑡, so that the phasor 𝑈 is 𝑈 = 5𝑒𝑗0° = 5∠0°.

The phasor form of the output is given by equations (10.41):

|𝑌| =𝑉𝑃

√1 + (𝜔𝑅𝐶)2=

5

√1 + (1000 ⋅ 1000 ⋅ 1 × 10−6)=

5

√2

∠𝜃𝑇 = − tan−1(𝜔𝑅𝐶) = − tan−1(1000 ⋅ 1000 ⋅ 1 × 10−6) = −𝜋

4= −45°

And the phasor 𝑌 can be written as 𝑌 =5

√2𝑒−𝑗45° =

5

√2∠ − 45°

We can create a phasor diagram of the input phasor 𝑈 and the output phasor 𝑌

Real

Imaginary

5

2

5

U

Y

45

The phasor diagram shows the input and output phasors in the complex plane. The magnitudes of the phasors are

typically labeled on the diagram, as is the phase difference between the two phasors. Note that since the phase

difference between 𝑌 and 𝑈 is negative, the output y(t) lags the input u(t).

The time-domain form of the output is:

𝑦(𝑡) =5

√2cos(1000𝑡 − 45°)

A time-domain plot of the input and output are shown below. This plot emphasizes that the output lags the input,

as indicated by our phasor diagram. The plot below replicates what would be seen from a measurement of the

input and output voltages.



Time, sec

1000

2

Voltage

Input, u(t)

Output, y(t)

0

45

Example 10.5: RL Circuit Sinusoidal Steady-state Response

In the circuit to the left below, the input voltage is 𝑉𝑃 cos(𝜔𝑡 + 30°) volts and the circuit response (or output) is

the inductor current, iL(t). We want to find the steady-state response 𝑖𝐿(𝑡 → ∞) .

Vpcos(t+30)+

-L

R

iL(t)+

-L

R

iL(t)Vpej(t+30)

The differential equation governing the circuit can be determined by applying KVL around the single loop:

𝐿𝑑𝑖𝐿(𝑡)

𝑑𝑡+ 𝑅𝑖𝐿(𝑡) = 𝑢(𝑡) Eq. 10.44

We apply the conceptual input, 𝑢(𝑡) = 𝑉𝑃𝑒𝑗(𝜔𝑡+30°) as shown in the figure to the right above to this equation. We

can represent this input in phasor form as:

𝑢(𝑡) = 𝑈𝑒𝑗𝜔𝑡 Eq. 10.45

Where the phasor 𝑈 = 𝑉𝑃∠30°. Likewise, we represent the output in phasor form:

𝑖𝐿(𝑡) = 𝐼𝐿𝑒𝑗𝜔𝑡 Eq. 10.46

Where the phasor 𝐼𝐿 = |𝐼𝐿|∠𝜃.

Substituting our assumed input and output in phasor form into equation (10.44) results in:

𝐿𝑗𝜔𝐼𝐿𝑒𝑗𝜔𝑡 + 𝑅𝐼𝐿𝑒𝑗𝜔𝑡 = 𝑈𝑒𝑗𝜔𝑡 Eq. 10.47

As in Example 10.4, we divide through by 𝑒𝑗𝜔𝑡 to obtain the frequency domain governing equation:

𝐿𝑗𝜔𝐼𝐿 + 𝑅𝐼𝐿 = 𝑈 Eq. 10.48

So that:

𝐼𝐿 =𝑈

𝑅+𝑗𝜔𝐿=

𝑉𝑃∠30°

𝑅+𝑗𝜔𝐿 Eq. 10.49

So that the phasor 𝐼𝐿 has magnitude and phase:



|𝐼𝐿| =𝑉𝑃

√𝑅2+(𝜔𝐿)2

𝜃 = 30° − tan−1 (𝜔𝐿

𝑅) Eq. 10.50

The exponential form of the inductor current is therefore:

𝑖𝐿(𝑡) =𝑉𝑃

√𝑅2+(𝜔𝐿)2𝑒𝑗[𝜔𝑡+30°−tan−1(

𝜔𝐿

𝑅)] Eq. 10.51

And the actual physical inductor current is:

𝑖𝐿(𝑡) =𝑉𝑃

√𝑅2+(𝜔𝐿)2cos [𝜔𝑡 + 30° − 𝑡𝑎𝑛−1 (

𝜔𝐿

𝑅)] Eq. 10.52

Example 10.6: Series RLC Circuit Sinusoidal Steady-state Response

Consider the circuit shown below. The input to the circuit is 𝑣𝑠(𝑡) = 2 cos(𝜔𝑡) volts. Find the output v(t).

+

-vs(t)

CR

v(t)+ -

L

In section 8.1, it was determined that the differential equation governing the system is:

𝑑2𝑣(𝑡)

𝑑𝑡2 +𝑅

𝐿

𝑑𝑣(𝑡)

𝑑𝑡+

1

𝐿𝐶𝑣(𝑡) =

1

𝐿𝐶𝑣𝑆(𝑡) Eq. 10.53

Assuming that the input is a complex exponential whose real part is the given vS(t) provides:

𝑣𝑆(𝑡) = 2𝑒𝑗𝜔𝑡 Eq. 10.54

The output is assumed to have the phasor form:

𝑣(𝑡) = 𝑉𝑒𝑗𝜔𝑡 Eq. 10.55

Where 𝑉 contains the (unknown) magnitude and phase of the output voltage. Substituting equations (10.54) and

(10.55) into equation (10.53) results in:

−(𝑗𝜔)2𝑉𝑒𝑗𝜔𝑡 +𝑅

𝐿(𝑗𝜔)𝑉𝑒𝑗𝜔𝑡 +

1

𝐿𝐶𝑉𝑒𝑗𝜔𝑡 =

1

𝐿𝐶2𝑒𝑗𝜔𝑡 Eq. 10.56

Dividing through by 𝑒𝑗𝜔𝑡 and noting that 𝑗2 = −1, results in:

[1

𝐿𝐶− 𝜔2 + 𝑗

𝑅

𝐿𝜔] 𝑉 =

2

𝐿𝐶

So that:

𝑉 =2

𝐿𝐶1

𝐿𝐶−𝜔2+𝑗

𝑅

𝐿𝜔

Eq. 10.57



The magnitude and phase of 𝑉 are:

|𝑉| =

2𝐿𝐶

√(1

𝐿𝐶− 𝜔2)

2

+ (𝑅𝐿

𝜔)2

∠𝑉 = − tan−1 (

𝑅𝜔𝐿

1𝐿𝐶

− 𝜔2)

And the capacitor voltage is:

𝑣(𝑡) =2

𝐿𝐶

√(1

𝐿𝐶−𝜔2)+(

𝑅

𝐿𝜔)

2cos 𝜔𝑡 − 𝑡𝑎𝑛−1 (

𝑅𝜔

𝐿1

𝐿𝐶−𝜔2

) Eq. 10.58

The complex arithmetic in this case becomes a bit tedious, but the complexity of the frequency-domain approach

is nowhere near that of the time-domain solution of the second-order differential equation.

Section Summary

• The steady-state response of a linear time invariant system to a sinusoidal input is a sinusoid with the

same frequency as the input sinusoid. Only the amplitude and phase angle of the output sinusoid can be

different from the input sinusoid, so the solution is entirely characterized by the magnitude and phase

angle of the output sinusoid. • The steady-state response of a system to a sinusoidal input can be determined by assuming a form of the

solution, substituting the input signal and the output signal into the governing differential equation and

solving for the amplitude and phase angle of the output sinusoid.

• The solution approach is simplified if the sinusoidal signals are represented as complex exponentials. The

approach is further simplified if these complex exponentials are represented in phasor form – the phasor

is a complex number which provides the amplitude and phase angle of the complex exponential.

• The above solution approaches convert the governing differential equation into an algebraic equation. If

complex exponentials in phasor form are used to represent the signals of interest, the governing algebraic

equation can have complex coefficients.

• The relationships between the steady state sinusoidal inputs and outputs are described by a relationship

between the amplitudes (generally a ratio between the output amplitude and the input amplitude) and

the phase angles (generally a difference between the output and input phase angles).

o These relationships are often displayed graphically in a phasor diagram.

10.3 Exercises

1. The differential equation governing a circuit is:

2𝑑𝑦(𝑡)

𝑑𝑡+ 6𝑦(𝑡) = 𝑢(𝑡)

Where u(t) is the input and y(t) is the output. Determine the steady-state response of the circuit to an

input 𝑢(𝑡) = 2 cos(3𝑡).

2. For the circuit shown below, u(t) is the input and y(t) is the output.

a. Write the differential equation relating u(t) and y(t).



b. Determine 𝑦(𝑡), 𝑡 → ∞, if 𝑢(𝑡) = 3cos (2𝑡)

u(t) 0.5F

+

-

y(t)+

-

1W

10.4 Phasor Representations of Circuit Elements

In section 10.3, we determined the sinusoidal steady-state response of an electrical circuit by transforming the

circuit’s governing differential equation into the frequency domain or phasor domain. This transformation

converted the differential equation into an algebraic equation. This conversion significantly simplified the

subsequent circuit analysis, at the relatively minor expense of performing some complex arithmetic. In this

module, we will further simplify this analysis by transforming the circuit itself directly into the frequency domain

and writing the governing algebraic equations directly.

This approach eliminates the necessity of ever writing the differential equation governing the circuit (as long as we

are only interested in the circuit’s sinusoidal steady-state response). This approach also allows us to apply analysis

techniques previously used only for purely resistive circuits to circuits containing energy storage elements.

10.4.1 Phasor Domain Voltage-current Relationships

In section 10.2, we introduced phasors as a method for representing sinusoidal signals. Phasors provide the

magnitude and phase of the sinusoid. For example, the signal 𝑣(𝑡) = 𝑉𝑃cos (𝜔𝑡 + 𝜃) has amplitude 𝑉𝑃 and the

phase angle 𝜃. This information can be represented in phasor form as:


In which complex exponentials are used to represent the phase. Equivalently, the phase can be represented as an

angle, and the phasor form of the signal can be written as:

𝑉 = 𝑉𝑃∠𝜃

Note that the phasor does not provide the frequency of the signal, ω. To include frequency information, the signal

is typically written in complex exponential form as:

𝑣(𝑡) = 𝑉𝑒𝑗𝜔𝑡

In section 10.3, we used phasor representations to determine the steady-state sinusoidal response of electrical

circuits by representing the signals of interest as complex exponentials in phasor form. When signals in the

governing differential equation are represented in this form, the differential equation becomes an algebraic

equation, resulting in a significant mathematical simplification. In section 10.3, it was also noted that the

mathematics could be simplified further by representing the circuit itself directly in the phasor domain. In this

section, we present the phasor form of voltage-current relations for our basic circuit elements: resistors, inductors,

and capacitors. The voltage-current relations for these elements are presented individually in the following sub-

sections.



10.4.2 Resistors

The voltage-current relationship for resistors is provided by Ohm’s Law:

𝑣(𝑡) = 𝑅 ⋅ 𝑖(𝑡) Eq. 10.59

If the voltage and current are represented in phasor form as:

𝑣(𝑡) = 𝑉𝑒𝑗𝜔𝑡 Eq. 10.60

And:

𝑖(𝑡) = 𝐼𝑒𝑗𝜔𝑡 Eq. 10.61

Equation (10.59) can be written:

𝑉𝑒𝑗𝜔𝑡 = 𝑅 ⋅ 𝐼𝑒𝑗𝜔𝑡 Eq. 10.62

Cancelling the 𝑒𝑗𝜔𝑡 term from both sides results in:

𝑉 = 𝑅 ⋅ 𝐼 Eq. 10.63

The voltage-current relationship for resistors (Ohm’s Law) is thus identical in the time and frequency domains.

Schematically, the time- and frequency-domain representations of a resistor are as shown in Fig. 10.10.

+

-

v(t)

i(t)

R

+

-

V

I

R

(a) Time domain (b) Frequency domain

Figure 10.10. Voltage-current relations for a resistor.

Equation (10.63) shows that, in the frequency domain, the voltage and current in a resistor are related by a purely

real, constant multiplicative factor. Thus, the sinusoidal voltage and current for a resistor are simply scaled

versions of one another – there is no phase difference in the voltage and current for a resistor. This is shown

graphically in Fig. 10.11.

Current

in phase

with

voltage

Current

Voltage

Time

v(t), i(t)

Figure 10.11. Voltage and current waveforms for a resistor.



A representative phasor diagram of the resistor’s voltage and current will appear as shown in Fig. 10.12 – the

phasors representing voltage and current will always be in the same direction, though their lengths will typically be

different.

Real

Imaginary

IRV =

I

Figure 10.12. Voltage-current phasor diagram for resistor.

10.4.3 Inductors

The voltage-current relationship for inductors is:

𝑣(𝑡) = 𝐿 ⋅𝑑𝑖(𝑡)

𝑑𝑡 Eq. 10.64

As with the resistive case presented above, we assume that the voltage and current are represented in phasor

form as 𝑣(𝑡) = 𝑉𝑒𝑗𝜔𝑡 and 𝑖(𝑡) = 𝐼𝑒𝑗𝜔𝑡, respectively. Substituting these expressions into equation (10.64) results

in:

𝑉𝑒𝑗𝜔𝑡 = 𝐿 ⋅𝑑

𝑑𝑡[𝐼𝑒𝑗𝜔𝑡] = 𝐿(𝑗𝜔)𝐼𝑒𝑗𝜔𝑡 Eq. 10.65

Dividing equation (10.65) by 𝑒𝑗𝜔𝑡 and re-arranging terms slightly results in the phasor domain or frequency

domain representation of the inductor’s voltage-current relationship:

𝑉 = 𝑗𝜔𝐿 ⋅ 𝐼 Eq. 10.66

In the frequency domain, therefore, the inductor’s phasor voltage is proportional to its phasor current. The

constant of proportionality is, unlike the case of the resistor, an imaginary number and is a function of the

frequency, ω. It is important to note that the differential relationship of equation (10.64) has been replaced with

an algebraic voltage-current relationship. Schematically, the time- and frequency-domain representations of an

inductor are as shown in Fig. 10.13.

+

-

vL(t)

iL(t)

L

+

-

LV

LI

Lj


Figure 10.13. Inductor voltage-current relations.



The factor of j in the voltage-current relationship of equation (10.66) introduces a 90∘ phase shift between

inductor voltage and current. Since 𝑗 = 𝑒𝑗90°, the voltage across an inductor leads the current by 90∘ (or,

equivalently, the current lags the voltage by 90∘). The relative phase difference between inductor voltage and

current are shown graphically in the time domain in Fig. 10.14. A representative phasor diagram of the inductor’s

voltage and current will appear as shown in Fig. 10.15 – the voltage phasor will always lead the current phasor

by 90∘, and the length of the voltage phasor will be a factor of ωL times the length of the current phasor.

Current

lags

voltage

by 90

Current

Voltage

Time

v(t), i(t)

Figure 10.14. Voltage and current waveforms for an inductor.

Real

Imaginary

IjV =

I

Figure 10.15. Voltage-current phasor diagram for inductor.

10.3.4 Capacitors

The voltage-current relationship for capacitors is:

𝑖(𝑡) = 𝐶 ⋅𝑑𝑣(𝑡)

𝑑𝑡 Eq. 10.67

As with the previous cases, we assume that the voltage and current are represented in phasor form as 𝑣(𝑡) =

𝑉𝑒𝑗𝜔𝑡 and 𝑖(𝑡) = 𝐼𝑒𝑗𝜔𝑡, respectively. Substituting these expressions into equation (10.67) results in:

𝐼𝑒𝑗𝜔𝑡 = 𝐶 ⋅𝑑

𝑑𝑡[𝑉𝑒𝑗𝜔𝑡] = 𝐶(𝑗𝜔)𝑉𝑒𝑗𝜔𝑡 Eq. 10.68

Dividing the above by 𝑒𝑗𝜔𝑡 results in the phasor domain or frequency domain representation of the capacitor's

voltage-current relationship:



𝐼 = 𝑗𝜔𝐶 ⋅ 𝑉 Eq. 10.69

To be consistent with our voltage-current relationship for resistors and capacitors, we write the voltage in terms of

the current. Thus,

𝑉 =1

𝑗𝜔𝐶⋅ 𝐼 Eq. 10.70

In the frequency domain, therefore, the capacitor’s phasor voltage is proportional to its phasor current. The

constant of proportionality is an imaginary number and is a function of the frequency, ω. As with inductors, the

differential voltage-current relationship has been replaced with an algebraic relationship. Schematically, the time-

and frequency-domain representations of a capacitor are as shown in Fig. 10.16.

+

-

vC(t)

iC(t)

C

+

-

CV

CI

Cj

1


Figure 10.16. Capacitor voltage-current relations.

The factor of 1

𝑗 in the voltage-current relationship of equation (10.70) introduces a 90∘ phase shift between

inductor voltage and current. Since 1

𝑗= 𝑒−𝑗90° = 1∠ − 90° , the voltage across a capacitor lags the current

by 90∘ (or, equivalently, the current leads the voltage by 90∘). The relative phase difference between capacitor

voltage and current are shown graphically in the time domain in Fig. 10.17. A representative phasor diagram of the

capacitor’s voltage and current will appear as shown in Fig. 10.18 – the voltage phasor will always lag the current

phasor by 90∘, and the length of the voltage phasor will be a factor of 1

𝜔𝐶 times the length of the current phasor.

Current

leads

voltage

by 90

Current

Voltage

Time

v(t), i(t)

Figure 10.17. Voltage and current waveforms for a capacitor.



Real

Imaginary

ICj

V

1=

I

Figure 10.18. Voltage-current phasor diagram for capacitor.

10.3.5 Impedance and Admittance

The frequency domain voltage-current characteristics presented in the previous subsections indicate that the

voltage difference across a circuit element can be written in terms of a multiplicative factor (which can be a

complex number) times the current through the element. In order to generalize and formalize this concept, we

define impedance as the ratio of phasor voltage to phasor current. Impedance is typically denoted as 𝑍 and is

defined mathematically as:

𝑍 =𝑉

𝐼 Eq. 10.71

Therefore, if the phasor voltage and current for a circuit element are given by:


And:

𝐼 = 𝐼𝑃𝑒𝑗𝜙

Then the impedance is:

𝑍 =𝑉

𝐼=

𝑉𝑃

𝐼𝑃𝑒𝑗(𝜃𝑍) Eq. 10.72

Or alternatively,

𝑍𝑉𝑃

𝐼𝑃∠𝜃𝑍 Eq. 10.73

Where 𝜃𝑍 is the angle of 𝑍. The magnitude of the impedance is the ratio of the magnitude of the voltage to the

magnitude of the current:

|𝑍| =𝑉𝑃

𝐼𝑃=

|𝑉|

|𝐼| Eq. 10.74

And the angle of the impedance is the difference between the voltage phase angle and the current phase angle:

𝜃𝑍 = ∠𝑍 = ∠𝑉 − ∠𝐼 = 𝜃 − 𝜙 Eq. 10.75

The impedance can also be represented in rectangular coordinates as:



𝑍 = 𝑅 + 𝑗𝑋 Eq. 10.76

Where R is the real part of the impedance (called the resistance or the resistive component of the impedance)

and X is the imaginary part of the impedance (called the reactance or the reactive part of the

impedance). R and X are related to |𝑍| and ∠𝑍 by the usual rules relating rectangular and polar coordinates, so

that:

|𝑍| = √𝑅2 + 𝑋2

∠𝑍 = tan−1 (𝑋

𝑅)

And:

𝑅 = 𝑅𝑒𝑍 = |𝑍| cos 𝜃𝑍

𝑋 = 𝐼𝑚𝑍 = |𝑍| sin 𝜃𝑍

Impedance is an extremely useful concept, in that it can be used to represent the voltage-current relations for any

two-terminal electrical circuit element in the frequency domain, as indicated in Fig. 10.19.

Electical

Circuit

+

-

I

VI

VZ =

Figure 10.19. Impedance representation of two-terminal electric circuit.

The admittance, 𝑌, is defined as the reciprocal of impedance:

𝑌 =1

𝑍 Eq. 10.77

Admittance is also a complex number, and is written in rectangular coordinates as:

𝑌 = 𝐺 + 𝑗𝐵 Eq. 10.78

Where G is called the conductance and B is the susceptance.

Impedances and admittances for the three electrical circuit elements presented previously in this section are

provided in Table 10.1 below. These results are readily obtained from the previously presented phasor domain

voltage-current relationships and the definitions of impedance and admittance. The relations provided in Table

10.1 should be committed to memory.



Element Impedance Admittance

Resistor

Inductor

Capacitor

R

Lj

Cj

1

Lj

1

R

1

Cj

Table 10.1. Impedances and admittances for passive circuit elements.

Example 10.7: Provide the Phasor-domain Representation of the Circuit Below

+

-20cos(10 t+30) 2W

0.1H

F30

1

The input amplitude is 20 volts, and the input phase is 30∘, so the phasor representation of the input voltage

is 20∠30∘.

The frequency of the input voltage is ω=10rad/sec. Thus, the impedances of the passive circuit elements are as

follows:

• Resistor: 𝑍 = 𝑅 = 2𝛺

• Inductor: 𝑍 = 𝑗𝜔𝐿 = 𝑗(10 𝑟𝑎𝑑 sec)(0.1𝐻) = 𝑗1𝛺⁄

• Capacitor: 𝑍 =1

𝑗𝜔𝐶=

1

𝑗(10𝑟𝑎𝑑 sec)(1

30𝐹)⁄

=3

𝑗Ω = −𝑗3Ω

The phasor-domain circuit is shown below.

+

-2W

j1W

-j3W3020

Section Summary

• Voltage-current relations for our passive circuit elements in the frequency domain are:

o Resistor: 𝑉 = 𝑅𝐼

o Inductor: 𝑉 = 𝑗𝜔𝐿 ⋅ 𝐼



o Capacitor: 𝑉 =1

𝑗𝜔𝐶⋅ 𝐼

• The impedance of a circuit element is the ratio of the phasor voltage to the phasor current in that

element:

o Resistor: 𝑍 = 𝑅

o Inductor: 𝑍 = 𝑗𝜔𝐿

o Capacitor: 𝑍 =1

𝑗𝜔𝐶

• Impedance is, in general, a complex number. Units of impedance are ohms (Ω). The real part of

impedance is the resistance. The imaginary part of impedance is reactance. Impedance, for general circuit

elements, plays the same role as resistance does for resistive circuit elements. In fact, for purely resistive

circuit elements, impedance is simply the resistance of the element.

• Admittance is the inverse of impedance.

• Admittance is, in general, a complex number. The real part of admittance is conductance. The imaginary

part of admittance is susceptance. For purely resistive circuits, admittance is the same as conductance.

• Impedance and admittance are, in general, functions of frequency.

• Impedance and admittance are not phasors. They are complex numbers – there is no sinusoidal time

domain function corresponding to impedance or admittance. (Phasors, by definition, are a way to

describe a time-domain sinusoidal function.)

10.4 Exercises

1. For the circuit shown below, u(t) is the input and y(t) is the output. Determine y(t), t→, if u(t) = 3cos(2t).

u(t) 0.5F

+

-

y(t)+

-

1W

2. Sketch a diagram of the input and output phasors for exercise 1 above.

3. Determine the impedance of the circuit elements shown below if Vin(t) = 2cos(4t).

F8

1

+

-

Vin(t) 2mH 5kW

+

-

Vin(t)

+

-

Vin(t)

4. Determine the impedance of the circuit elements in exercise 3 if Vin(t) = 3cos(8t)

10.5 Direct Frequency Domain Circuit Analysis

In section 10.3, we determined the steady-state response of electrical circuits to sinusoidal signals using phasor

representations of the signals involved, and time-domain representations of the circuit element voltage-current

relations. Applying KVL and KCL in this manner resulted in governing equations in which the time dependence had

been removed, which converted the governing equations from differential equations to algebraic equations.



Unknowns in the resulting algebraic equations were the phasor representations of the signals. These equations

could then be solved to determine the desired signals in phasor form; these results could then be used to

determine the time-domain representations of the signals.

In section 10.4, we replaced the time-domain voltage-current relations for passive electrical circuit elements with

impedances, which provide voltage-current relations for the circuit elements directly in the frequency domain. At

the end of section 10.4, we used these impedances to schematically represent a circuit directly in the frequency

domain.

In this section, we will use this frequency-domain circuit representation to perform circuit analysis directly in the

frequency domain using phasor representations of the signals and impedance representations of the circuit

elements. This will allow us to write the algebraic equations governing the phasor representation of the

circuit directly, without any reference to the time domain behavior of the circuit. As in section 10.3, these

equations can be solved to determine the behavior of the circuit in terms of phasors, and the results transformed

to the time domain.

Performing the circuit analysis directly in the frequency domain using impedances to represent the circuit

elements can result in a significant simplification of the analysis. In addition, many circuit analysis techniques

which were previously applied to resistive circuits (e.g. circuit reduction, nodal analysis, mesh analysis,

superposition, Thevenin’s and Norton’s Theorems) are directly applicable in the frequency domain. Since these

analysis techniques have been presented earlier for resistive circuits, in this section we will simply:

1. Provide examples of applying these analysis methods to frequency-domain circuits, and

2. Note any generalizations relative to using phasors in these analysis methods

Throughout this section, the reader should firmly keep in mind that we are dealing only with the steady-state

responses of circuits to sinusoidal forcing functions. It is sometimes easy to lose track of this fact, since the

sinusoidal nature of the signal is often not explicitly stated, but any time we deal with impedances and phasors, we

are working with sinusoidal signals.

10.5.1 Kirchhoff’s Voltage Law

Kirchhoff’s Voltage Law states that the sum of the voltage differences around any closed loop is zero. Therefore,

if v1(t), v2(t), …, vN(t) are the voltages around some closed loop, KVL provides:

∑ 𝑣𝑘(𝑡) = 0𝑁𝑘=1 Eq. 10.79

Substituting the phasor representation of the voltages results in:

∑ 𝑉𝑘𝑒𝑗𝜔𝑡 = 0𝑁𝑘=1 Eq. 10.80

Dividing equation (10.80) by 𝑒𝑗𝜔𝑡 results in:

∑ 𝑉𝑘 = 0𝑁𝑘=1 Eq. 10.81

So that KVL states that the sum of the phasor voltages around any closed loop is zero.

10.5.2 Kirchhoff's Current Law

Kirchhoff's Current Law states that the sum of the currents entering any node is zero. Therefore,

if 𝑖1(𝑡), 𝑖2(𝑡), ⋯ , 𝑖𝑁(𝑡) are the currents entering a node, KCL provides:

∑ 𝑖𝑘(𝑡) = 0𝑁𝑘=1 Eq. 10.82



Substituting the phasor representation of the currents results in:

∑ 𝐼𝑘𝑒𝑗𝜔𝑡𝑁𝑘=1 = 0 Eq. 10.83

Dividing equation (10.83) by 𝑒𝑗𝜔𝑡 results in:

∑ 𝐼𝑘 = 0𝑁𝑘=1 Eq. 10.84

So that KCL states that the sum of the phasor currents entering (or leaving) a node is zero.

Important Result: KVL and KCL apply directly in the frequency domain.

Example 10.8: RC Circuit Steady-state Sinusoidal Response

In this example, we will revisit example 10.3. In that example, we determined the capacitor voltage in the circuit to

the left below, using phasor analysis techniques applied to the circuit’s time-domain governing equation. In this

example, we will represent the circuit itself directly in the frequency domain, using impedance representations of

the circuit element. The frequency-domain representation of the circuit is shown to the right below.

+

-Vpcos(t) C

R

y(t)

+

-

+

-

R +

-

0PVCj

1Y

I

By the definition of impedance, we can determine the current through the capacitor to be:

𝐼 =𝑌

𝑍𝐶

=𝑌

1𝑗𝜔𝐶

= 𝑗𝜔𝐶𝑌

The voltage across the resistor can now, by the definition of impedance, be written as 𝑉𝑅 = 𝑅 ⋅ 𝐼 = 𝑅(𝑗𝜔𝐶𝑌). We

now apply KVL for phasors to the circuit to the right above, which leads to:

𝑉𝑃∠0° = 𝑅(𝑗𝜔𝐶𝑌) + 𝑌

Solving for 𝑌 in this equation provides 𝑌 =𝑉𝑃∠0°

1+𝑗𝜔𝑅𝐶

By the rules of complex arithmetic, we can determine the magnitude and phase angle of 𝑌 to be:

|𝑌| =𝑉𝑃

√1 + (𝜔𝑅𝐶)2

∠𝑌 = − tan−1(𝜔𝑅𝐶)

And the time-domain solution for y(t) is thus:

𝑦(𝑡) =𝑉𝑃

√1 + (𝜔𝑅𝐶)2cos[𝜔𝑡 − tan−1(𝜔𝑅𝐶)]



10.5.3 Parallel and Series Impedances & Circuit Reduction

Consider the case of N impedances connected in series, as shown in Fig. 10.20. Since the elements are in series,

and since we have seen that KCL applies to phasors, the phasor current 𝐼 flows through each of the impedances.

Applying KVL for phasors around the single loop, and incorporating the definition of impedance, we obtain:

𝑉 = 𝐼(𝑍1 + 𝑍2 + ⋯ + 𝑍𝑁) = 0 Eq. 10.85

+ -1V

1Z

+ -2V

2Z+

-

NVNZ

I

V

+

-

Figure 10.20. Series combination of impedances.

If we define 𝑍𝑒𝑞 as the equivalent impedance of the series combination, we have 𝑉 = 𝐼 ⋅ 𝑍𝑒𝑞 , where:

𝑍𝑒𝑞 = 𝑍1 + 𝑍2 + ⋯ + 𝑍𝑁 Eq. 10.86

So that impedances in series sum directly. Thus, impedances in series can be combined in the same way as

resistances in series.

By extension of the above result, we can develop a voltage divider formula for phasors. Without derivation, we

state that the phasor voltage across the kth impedance in a series combination of N impedances as shown in Fig.

10.20 can be determined as:

𝑉𝑘 = 𝑉(𝑌1 + 𝑌2 + ⋯ + 𝑌𝑁) = 0 Eq. 10.88

I

V

+

-

1I1Z 2I

2Z NINZ

Figure 10.21. Parallel combination of impedances.

If we define 𝑌𝑒𝑞 as the equivalent impedance of the series combination, we have:

𝐼 = 𝑉 ⋅ 𝑌𝑒𝑞 Eq. 10.89

Where:

𝑌𝑒𝑞 = 𝑌1 + 𝑌2 + ⋯ + 𝑌𝑁 Eq. 10.90

So that admittances in series sum directly. Converting our admittances to impedances indicates that the

equivalent impedance of a parallel combination of N impedances as shown in Fig. 10.21 is:



𝑍𝑒𝑞 =1

1

𝑍1+

1

𝑍2+⋯+

1

𝑍𝑁

Eq. 10.91

Thus, impedances in parallel can be combined in the same way as resistances in parallel.

By extension of the above result, we can develop a current divider formula for phasors. Without derivation, we

state that the phasor current across the kth impedance in a series combination of N impedances as shown in Fig.

10.21 can be determined as:

𝐼𝑘 = 𝐼

1

𝑍𝑘1

𝑍1+

1

𝑍2+⋯+

1

𝑍𝑁

Eq. 10.92

So that our current division relationships for resistors in parallel apply directly in the frequency domain for

impedances in parallel.

Important Result: All circuit reduction techniques for resistances apply directly to the frequency domain for

impedances. Likewise, voltage and current divider relationships apply to phasor circuits in the frequency domain

exactly as they apply to resistive circuits in the time domain.

Example 10.9

Use circuit reduction techniques to determine the current phasor 𝐼 leaving the source in the circuit below. (Note:

the circuit below is the frequency domain circuit we obtained in example 10.7.)

+

-2W

j1W

-j3W3020

I

Since impedances in series add directly, the inductor and resistor can be combined into a single equivalent

impedance of (2 + 𝑗1)𝛺, as shown in the figure to the left below. The capacitor is then in parallel with this

equivalent impedance. Since impedances in parallel add in the same way as resistors in parallel, the equivalent

impedance of this parallel combination can be calculated by dividing the product of the impedances by their sum,

so 𝑍𝑒𝑞 =(−𝑗3)(2+𝑗1)

(−𝑗3)+(2+𝑗1)Ω =

3−𝑗6

2−𝑗2Ω. Converting this impedance to polar form results in 𝑍𝑒𝑞 = 2.37∠ − 18°Ω; the final

reduced circuit is shown in the figure to the right below.

+

-(2+j1)W-j3W3020

I

+

-3020

I

W− 18372.

Using the reduced circuit to the right above and the definition of impedance, we can see that:



𝐼 =20∠30°

2.37∠ − 18°Ω=

2

2.37∠[30° − (−18°)]𝐴

So that:

𝐼 = 8.44∠48°𝐴

Example 10.10

Use circuit reduction techniques to determine the current, i(t) through the inductor in the circuit below.

+

-

2W

5cos(2t)

4W

0.125F

2H

i(t)

With =2 rad/sec, the frequency domain representation of the circuit is as shown in the figure to the left below; in

that figure, we have also defined the current phasor 𝐼𝑆 leaving the source.

j4W

-j4W

I

+

-

2W

4W

05

SI

We now employ circuit reduction techniques to determine the phasor 𝐼. To do this, we first determine the circuit

impedance seen by the source; this impedance allows us to determine the source current 𝐼𝑆. The current 𝐼 can be

determined from a current divider relation and 𝐼𝑆.

The impedances of the series combination of the capacitor and the 4W resistor is readily obtained by adding their

individual impedances, as shown in the figure to the left below. This equivalent impedance is then in parallel with

the inductor’s impedance; the equivalent impedance of this parallel combination is as shown in the circuit to the

right below.

j4W(4-j4)W

I

+

-

2W

05

SI

(4-j4)W+

-

2W

05

SI

The source current is then, by the definition of impedance, 𝐼𝑆 =5∠0°

(4−𝑗4)Ω+2Ω= 0.69∠ − 33.7°.

The circuit to the left above, along with our voltage divider formula, provides:



𝐼 =(4 − 𝑗4)Ω

(4 − 𝑗4)Ω + 𝑗4Ω⋅ 𝐼𝑆 = (1 − 𝑗1)Ω ⋅ 0.69∠ − 33.7° = 0.98∠ − 78.7°

And the current 𝑖(𝑡) = 0.98 cos(2𝑡 − 78.7°)

10.5.4 Nodal and Mesh Analysis

Nodal analysis and mesh analysis techniques have been previously applied to resistive circuits in the time domain.

In nodal analysis, we applied KCL at independent nodes and used Ohm’s Law to write the resulting equations in

terms of the node voltages. In mesh analysis, we applied KVL and used Ohm’s Law to write the resulting equations

in terms of the mesh currents.

In the frequency domain, as we have seen in previous sub-sections, KVL and KCL apply directly to the phasor

representations of voltages and currents. Also, in the frequency domain, impedances can be used to represent

voltage-current relations for circuit elements in the frequency domain in the same way that Ohm’s Law applied to

resistors in the time domain (the relation 𝑉 = 𝐼 ⋅ 𝑍 in the frequency domain corresponds exactly to the

relation 𝑣(𝑡) = 𝑅 ⋅ 𝑖(𝑡) in the time domain). Thus, nodal analysis and mesh analysis apply to frequency domain

circuits in exactly the same way as to time domain resistive circuits, with the following modifications:

• The circuit excitations and responses are represented by phasors

• Phasor representations of node voltages and mesh currents are used

• Impedances are used in the place of resistances

Application of nodal and mesh analysis to frequency-domain circuit analysis is illustrated in the following

examples.

Example 10.11

Use nodal analysis to determine the current i(t) in the circuit of example 10.10.

The desired frequency-domain circuit was previously determined in Example 10.10. Nodal analysis of the

frequency-domain circuit proceeds exactly as was done in the case of resistive circuits. The reference voltage, 𝑉𝑅 =

0, and our single node voltage, VA, for this circuit are defined on the circuit below.

j4W

-j4W

I

+

-

2W

4W

05

VA

VR=0

Applying KCL in phasor form at node A provides:

5∠0° − 𝑉𝐴

2Ω−

𝑉𝐴

(4 − 𝑗4)Ω−

𝑉𝐴

𝑗4Ω= 0



Solving for 𝑉𝐴 gives 𝑉𝐴 = 3.92∠11.31°𝑉. By the definition of impedance, the desired current phasor 𝐼 =𝑉𝐴

𝑗4Ω=

3.92∠11.31°

4∠90°= 0.98∠ − 78.7° so that 𝑖(𝑡) = 0.98cos (2𝑡 − 78.7°), which is consistent with our result obtained via

circuit reduction in Example 3.

Example 10.12

Use mesh analysis to determine the current i(t) in the circuit of examples 10.10 and 10.11.

The desired frequency-domain circuit was previously determined in Example 10.10. Mesh analysis of the

frequency-domain circuit proceeds exactly as for resistive circuits. The figure below shows our choice of mesh

loops; the series resistor-capacitor combination has been combined into a single equivalent resistance in the figure

below, for clarity.

+

-

2W

j4W

I

05 1I 2I(4-j4)W

KVLO around the mesh loop 𝐼1 provides:

5∠0° − 2 ⋅ 𝐼1 − (4 − 𝑗4)(𝐼1 − 𝐼2) = 0

KVL around the mesh loop 𝐼2 provides:

(4 − 𝑗4)(𝐼2 − 𝐼1) + 𝑗4 ⋅ 𝐼2 = 0

The second equation above can be simplified to provide: 𝐼2 = (1 − 𝑗)𝐼1. Using this result to eliminate 𝐼1 in the

mesh equation for loop 𝐼1 and simplifying provides:

5∠0° = [(6 − 𝑗4)

1 − 𝑗+ (𝑗4 − 4)] 𝐼2

So that 𝐼2 = 0.98∠ − 78.7°. The mesh current 𝐼2 is simply the desired current 𝐼, so in the time domain:

𝑖(𝑡) = 0.98 cos(2𝑡 − 78.7°)

Which is consistent without results from examples 10.10 and 10.11.

Important Result: Nodal and mesh analysis methods apply to phasor circuits exactly as they apply to resistive

circuits in the time domain. Impedances simply replace resistances, and quantities of interest become complex

valued.

10.5.5 Superposition

The extension of superposition to the frequency domain is an extremely important topic. Several common analysis

techniques you will encounter later in this course and in future courses (frequency response, Fourier Series, and



Fourier Transforms, for example) will depend heavily upon the superposition of sinusoidal signals. In this sub-

section, we introduce the basic concepts involved.

In all of our steady-state sinusoidal analyses, we have required that the circuit is linear. (The statement that the

steady state response to a sinusoidal input is a sinusoid at the same frequency requires the system to be linear.

Nonlinear systems do not necessarily have this characteristic.) Thus, all phasor circuits are linear and superposition

must apply. Thus, if a phasor circuit has multiple inputs, we can calculate the response of the circuit to each input

individually and sum the results to obtain the overall response. It is important to realize, however, that the final

step of summing the individual contributions to obtain the overall response can, in general, only be done in the

time domain. Since the phasor representation of the circuit response implicitly assumes a particular frequency, the

phasor representations cannot be summed directly. The time domain circuit response, however, explicitly provides

frequency information, allowing those responses to be summed.

In fact, because the frequency-domain representation of the circuit depends upon the frequency of the input (in

general, the impedances will be a function of frequency), the frequency domain representation of the circuit itself

is, in general, different for different inputs. Thus, the only way in which circuits with multiple inputs at different

frequencies can be analyzed in the frequency domain is with superposition.

In the special case in which all inputs share a common frequency, the circuit response can be determined by any of

our previous analysis techniques (circuit reduction, nodal analysis, mesh analysis, superposition, etc.) In this case, if

superposition is used, the circuit response to individual inputs can be summed directly in the frequency domain if

desired.

Examples of the application of superposition to analysis of frequency-domain circuits are provided below.

Important Result: In the case of multiple frequencies existing in the circuit, superposition is the only valid

frequency-domain analysis approach.

Superposition applies directly in the frequency domain, insofar as contributions from individual sources can be

determined by killing all other sources and analyzing the resulting circuit. In general, however, superimposing

(summing) the contributions from the individual sources must be done in the time domain.

Superposition of responses to individual sources can be summed directly in the frequency domain (e.g. addition of

the phasors representing the individual responses) is only appropriate if all sources have the same frequency. In

this case (all source having the same frequency) any of our other modeling approaches are also valid.

Example 10.13

Determine the voltage v(t) across the inductor in the circuit below.

6cos(9t) A

3W1W

4cos(3t+30) VH3

1 +

-v(t)

+

-

Since two different input frequencies are applied to the circuit, we must use superposition to determine the

response. The circuit to the left below will provide the phasor response 𝑉1 to the current source; the frequency is



= 9 rad/sec and the voltage source is killed. The circuit to the right below will provide the phasor response 𝑉2 to

the voltage source; the frequency is = 3 rad/sec and the current source is killed.

3W1W

+

-

j3W06

3W1W

+

-

j1W 304+

-1V 2V

To determine the voltage phasor resulting from the current source (𝑉 in the circuit to the left above), we note that

the inductor and the 3Ω resistor form a current divider. Thus, the current through the inductor resulting from the

current source is 𝐼1 =3Ω

(3+𝑗3)Ω⋅ 6∠0° =

3∠0°⋅∠0°

3√2∠45°=

6

√2∠ − 45°. The voltage phasor 𝑉 can then be determined by

multiplying this current times the inductor’s impedance:

𝑉1 = 𝑗3Ω ⋅6

√2∠ − 45° = 3∠90° ⋅

6

√2∠ − 45°𝑉

And the time-domain voltage across the inductor due to the current source is:

𝑣1(𝑡) = 9√2 cos(9𝑡 + 45°)𝑉

To determine the voltage phasor resulting from the voltage source (𝑉2 in the circuit to the right above), we note

that the inductor and the 3Ω resistor now form a voltage divider. Thus, the voltage 𝑉2 can be readily determined

by:

𝑉2 =𝑗1Ω

(3 + 𝑗1)Ω⋅ 4∠30° =

1∠90° ⋅ 4∠30°

√10∠18.4°=

4

√10∠101.6°

So that the time-domain voltage across the inductor due to the voltage source is:

𝑣2(𝑡) =4

√10cos(3𝑡 + 101.6°)𝑉

The overall voltage is then the sum of the contributions from the two sources, in the time domain, so:

𝑣(𝑡) = 𝑣1(𝑡) + 𝑣2(𝑡)

And:

𝑣(𝑡) = 9√2 cos(9𝑡 + 45°) +4

√10cos(3𝑡 + 101.6°)𝑉

Example 10.14

Determine the voltage v(t) across the inductor in the circuit below.



6cos(9t) A

3W1W

4cos(9t+30) VH3

1 +

-v(t)

+

-

This circuit is essentially the same as the circuit of Example 10.13, with the important difference that the frequency

of the voltage input has changed – the voltage source and current source both provide the same frequency input

to the circuit, 9 rad/sec. We will first do this problem using superposition techniques. We will then use nodal

analysis to solve the problem, to illustrate that multiple inputs at the same frequency do not require the use of

superposition.

Individually killing each source in the circuit above results in the two circuits shown below. Note that the

impedance of the inductor is now the same in both of these circuits.

3W1W

+

-

j3W06

3W1W

+

-

j3W 304+

-1V 2V

The two circuits shown above will now be analyzed to determine the individual contributions to the inductor

voltage; these results will then be summed to determine the overall inductor voltage.

The circuit to the left above has been analyzed in Example 6. Therefore, the voltage phasor 𝑉1 is the same as

determined in Example 10.13:

𝑉1 = 9√2∠45°𝑉

The voltage 𝑉2 in the circuit to the right above can be determined from application of the voltage divider formula

for phasors:

𝑉2 =𝑗3Ω

(3 + 𝑗3)Ω⋅ 4∠30° =

3∠90° ⋅ 4∠30°

3√2∠45°= 2√2∠75°

Since both inputs have the same frequency, we can superimpose the phasor results directly (we could, of course,

also determine the individual time domain responses and superimpose those responses if we chose):

𝑉 = 𝑉1 + 𝑉2 = 9√2∠45° + 2√2∠75° = 15.24∠50.3°𝑉

So that the time domain inductor voltage is 𝑣(𝑡) = 15.24 cos(9𝑡 + 50.3°)𝑉. Notice that the circuit response has

only a single frequency component, since both inputs have the same frequency.

The superposition approach provided above is entirely valid. However, since both sources have the same input, we

can choose any of our other analysis approaches to perform this problem. To emphasize this fact, we choose to do

this problem using nodal analysis.



The frequency-domain circuit, with our definition of reference voltage and independent node, is shown in the

figure below.

3W1W

+

-

+

-

06 304V

VA

VR=0

j3W

KCL at node A provides:

6∠0° =𝑉𝐴 − 0

𝑗3Ω+

𝑉𝐴 − 4∠30°

3Ω

Solving the above equation for 𝑉𝐴 provides 𝑉𝐴 = 15.24∠50.3𝑉 so that the inductor voltage as a function of time is:

𝑣(𝑡) = 15.24 cos(9𝑡 + 50.3°)𝑉

Which is consistent with our result using superposition.

10.5.6 Thévenin’s & Norton’s Theorems, Source Transformations, and Maximum

Power Transfer

Application of Thévenin’s and Norton’s Theorems to frequency domain circuits is identical to their application to

time domain resistive circuits. The only differences are:

• The open circuit voltage (VOC) and short circuit current (iSC) determined for resistive circuits is replaced by

their phasor representations, 𝑉𝑂𝐶 and 𝐼𝑆𝐶 .

• The Thévenin resistance, RTH, is replaced by a Thévenin impedance, 𝑍𝑇𝐻.

Thus, the Thévenin and Norton equivalent circuits in the frequency domain are as shown in Fig. 10.22.

THZ

+

-OCVTHZ

SCI

(a) Thévenin circuit (b) Norton circuit

Figure 10.22. Thévenin and Norton equivalent circuits.

Since Thévenin’s and Norton’s Theorems both apply in the frequency domain, the approaches we used for source

transformations in the time domain for resistive circuits translate directly to the frequency domain, with

impedances substituted for resistances and phasors used for voltage and current terms.



In order to determine the load necessary to draw the maximum power from a Thévenin equivalent circuit, we

must re-derive the maximum power result obtained previously for resistive circuits, substituting impedances for

admittances and using phasors for source terms. We will not derive the governing relationship, but will simply

state that, in order to transfer the maximum power to a load, the load impedance must be the complex conjugate

of the Thévenin impedance of the circuit being loaded. Thus, if a Thévenin equivalent circuit has some

impedance 𝑍𝑇𝐻 with a resistance RTH and a reactance XTH, the load which will draw the maximum power from this

circuit must have resistance RTH and a reactance – XTH. The appropriate loaded circuit is shown in Fig. 10.23 below.

THThTH jXRZ +=

+-

OCV THThL jXRZ −=

Figure 10.23. Load impedance to draw maximum power from a Thévenin circuit.

Example 10.15

Determine the Thévenin equivalent circuit seen by the load in the circuit below

+

-2cos(2t)

2W

2W

0.5H

Load

In the circuit below, we have used the input frequency, ω=2 rad/sec, to convert the circuit to the frequency

domain.

02+

-

2W

2W

j1W

LZ

Removing the load and killing the source allows us to determine the Thévenin resistance of the circuit. The

appropriate circuit is:



2W

2W

j1W

THZ

The parallel combination of two, 2W resistors have an equivalent resistance of 1W. This impedance, in series with

the j1W impedance, results in a Thévenin impedance 𝑍𝑇𝐻 = (1 + 𝑗1)Ω. Replacing the source, but leaving the load

terminals open-circuited, as shown in the figure below, allows us to determine the open-circuit voltage 𝑉𝑂𝐶 .

02+

-

2W

2W

j1W

OCV

+

-

Since there is no current through the inductor, due to the open-circuit condition, 𝑉𝑂𝐶 is determined from a simple

resistive voltage divider formed by the two, 2Ω resistors. Thus, the open-circuit voltage is:

𝑉𝑂𝐶 =2Ω

2Ω + 2Ω⋅ 2∠0° = 1∠0°

The resulting Thévenin equivalent circuit is shown below:

+

-

01

(1+j1)W

Example 10.16

Determine the Norton equivalent circuit of the circuit of example 10.15.

Since we determined the Thévenin equivalent circuit in Example 10.15, a source transformation can be used to

determine the Norton equivalent circuit. Consistent with our previous source transformation rules, the short-

circuit current, 𝐼𝑆𝐶 , is equal to the open-circuit voltage divided by the Thévenin impedance:

𝐼𝑆𝐶 =𝑉𝑂𝐶

𝑍𝑇𝐻

=1∠0°

(1 + 𝑗1)Ω=

1∠0°

√2∠45°=

1

√2∠ − 45°

Since the impedance doesn’t change during a source transformation, the Norton equivalent circuit is therefore as

shown below:



+

-

01

(1+j1)W

Example 10.17

Determine the load impedance for the circuit of Example 10.15 which will provide the maximum amount of power

to be delivered to the load. Provide a physical realization (a circuit) which will provide this impedance.

The maximum power is delivered to the load when the load impedance is the complex conjugate of the Thévenin

impedance. Thus, the load impedance for maximum power transfer is:

𝑍𝐿 = (1 − 𝑗)Ω

And the loaded Thévenin circuit is:

+

-01

(1+j1)W

(1-j1)W

To implement this load, let us look at a parallel RC combination. With the frequency ω=2 rad/sec, the frequency

domain load looks like:

RC

j

2−

LZ

Combining parallel impedances results in:

𝑍𝐿 =−𝑗 (

𝑅2𝐶

)

𝑅 −𝑗

2𝐶

Ω =

𝑅4𝐶2 − 𝑗

𝑅2

2𝐶

𝑅2 +1

4𝐶2

Ω

Setting 𝑅 = 2Ω and 𝐶 = 0.25𝐹 makes 𝑍𝐿 = (1 − 𝑗)Ω, as desired, so the physical implementation of our load is as

shown below:



2W 0.25F

Section Summary

• The following analysis methods apply in the frequency domain exactly as they do in the time domain for

purely resistive circuits o KVL and KCL

o Voltage and current dividers

o Circuit reduction techniques

o Nodal and mesh analysis

o Superposition, especially when multiple frequencies are present

o Thévenin’s and Norton’s theorems

One simply uses phasor representations for the voltages and/or currents in the circuit and impedances to

represent the circuit element voltage-current relationships. The analysis techniques presented Chapters 1 through

4 are then applied exactly as they were for resistive circuits.

• One minor exception to the above statement is that, in order to draw maximum power from a circuit, the

load impedance should be the complex conjugate of the impedance of the circuit’s Thévenin equivalent.

10.5 Exercises:

1. Determine the impedance seen by the source for the circuit below if u(t) = 4cos(t+30).

u(t)+

-

2W

F4

14W

2. Determine the impedance seen by the source for the circuit below if u(t) = 2cos(4000t).

u(t)+

-8W 8W

1mH

3. For the circuit of exercise 1, determine the current delivered by the source.

4. For the circuit of exercise 2, determine the voltage across the 8W resistors.



10.6 Frequency Domain System Characterization

In Chapters 7 and 8, we wrote the differential equation governing the relationship between a circuit’s input and

output (the input-output equation) and used this differential equation to determine the response of a circuit to

some input. We also characterized the time-domain behavior of the system by examining the circuit’s natural and

step responses. We saw that the behavior of a first order circuit can be characterized by its time constant and DC

gain, while the response of a second order circuit is characterized by its natural frequency, damping ratio and DC

gain. It is important to recognize that these characterizations were independent of specific input parameters; they

depended upon the type of response (e.g. a step function or a natural response), but were independent of detailed

information such as the amplitude of the step input or the actual values of the initial conditions.

We will now use the steady state sinusoidal response to characterize a circuit’s behavior. As in the case of our

time-domain characterization, this characterization will allow the system’s behavior to be defined in terms of its

response to sinusoidal inputs, but the characterization will be independent of details such as the input sinusoid’s

amplitude or phase angle. (The input sinusoid’s frequency will, however, still be of prime importance.)

When a sinusoidal input is applied to a linear system, the system’s forced response consists of a sinusoid with the

same frequency as the input sinusoid, but in general having a different amplitude and phase from the input

sinusoid. Figure 10.24 shows the general behavior, in block diagram form. Changes in the amplitude and phase

angle between the input and output signals are often used to characterize the circuit’s input-output relationship at

the input frequency, ω. In this chapter, we will demonstrate how this characterization is performed for inputs with

discrete frequencies (as in the case of circuits with one or several inputs, each with a single frequency component).

Later chapters will extend these concepts to the case in which frequency is considered to be a continuous variable.

SystemInput

u(t)=Acos(t+)

Output

y(t)=Bcos(t+f)

Figure 10.24. Sinusoidal steady-state input-output relation for a linear time invariant system.

Previously in this chapter, we have (bit-by-bit) simplified the analysis of a system’s steady state sinusoidal response

significantly. We first represented the sinusoidal signals as complex exponentials in order to facilitate our analysis.

We subsequently used phasors to represent our complex exponential signals, as shown in Fig. 10.25; this allowed

us to represent and analyze the circuit’s steady state sinusoidal response directly in the frequency domain.

Input OutputSystem

= AU f= BY

Figure 10.25. Phasor representation of sinusoidal inputs and outputs.

In the frequency domain analyses performed to date, we have generally determined the system’s response to a

specific input signal with a given frequency, amplitude, and phase angle. We now wish to characterize the system

response to an input signal with a given frequency, but an arbitrary amplitude and phase angle. As indicated

previously in section 10.1, we will see that the input-output relationship governing the system reduces to a

relationship between the output and input signal amplitudes and the output and input signal phases. The circuit

can thus be represented in phasor form as shown in Fig. 10.26. The system’s effect on a sinusoidal input consists of

an amplitude gain between the output and input signals (𝐵

𝐴 in Fig. 10.26) and a phase difference between the

output and input signals (𝜙 − 𝜃 in Fig. 10.26).



Input Output( )f −

A

B= AU f= BY

Figure 10.26. Frequency domain representation of circuit input-output relationship.

Rather than perform a rigorous demonstration of this property at this time, we will simply provide some simple

examples to illustrate the basic concept.

Example 10.18

A sinusoidal voltage, vin(t), is applied to the circuit to the left below. Determine the frequency-domain relationship

between the phasor representing vin(t) and the phasor representing the output voltage vout(t).

+

-C

R +

-

vin(t) vout(t)

+

-

R +

-

inVoutV

Cj

1

Since the frequency is unspecified, we leave frequency as an independent variable, , in our analysis. In the

frequency domain, therefore, the circuit can be represented as shown to the right above. The frequency domain

circuit is a simple voltage divider, so the relation between input and output is:

𝑉𝑜𝑢𝑡 =

1𝑗𝜔𝐶

𝑅 +1

𝑗𝜔𝐶

⋅ 𝑉𝑖𝑛 =1

1 + 𝑗𝜔𝑅𝐶⋅ 𝑉𝑖𝑛

The factor 1

1+𝑗𝜔𝑅𝐶 is a complex number, for given values of , R, and C. It constitutes a multiplicative factor which,

when applied to the input, results in the output. This multiplicative factor is often used to characterize the

system’s response at some frequency, . We will call this multiplicative factor the frequency response function,

and denote it as H(j). For a particular frequency, H(j) is a complex number, with some amplitude, |𝐻(𝑗𝜔)|, and

phase angle, ∠𝐻(𝑗𝜔). For our example, the magnitude and phase of our frequency response function are:

|𝐻(𝑗𝜔)| =1

√1 + (𝜔𝑅𝐶)2

∠𝐻(𝑗𝜔) = − tan−1(𝜔𝑅𝐶)

According to the rules of multiplication of complex numbers, when two complex numbers are multiplied, the

magnitude of the result is the product of the magnitudes of the individual numbers, and the phase angle of the

result is the sum of the individual phase angles. Thus, if the input voltage is represented in phasor form as 𝑉𝑖𝑛 =

|𝑉𝑖𝑛| ⋅ |𝐻(𝑗𝜔)| and the output voltage is 𝑉𝑜𝑢𝑡 = |𝑉𝑜𝑢𝑡|∠𝜙, it is easy to obtain the output voltage from the input

voltage and the frequency response function:

|𝑉𝑜𝑢𝑡| = |𝑉𝑖𝑛| ⋅ |𝐻(𝑗𝜔)|

∠𝑉𝑜𝑢𝑡 = ∠𝑉𝑖𝑛 + ∠𝐻(𝑗𝜔)



Example 10.19

Use the frequency response function determined in Example 10.18 above to determine the response vout(t) of the

circuit shown below to the following input voltages:

• 𝑣𝑖𝑛(𝑡) = 3 cos(2𝑡 + 20°)

• 𝑣𝑖𝑛(𝑡) = 7 cos(4𝑡 − 60°)

+

-0.25F

2W +

-

vin(t) vout(t)

When 𝑣𝑖𝑛(𝑡) = 3 cos(2𝑡 + 20°), 𝜔 = 2 rad/sec, |𝑉𝑖𝑛| = 3 and ∠𝑉𝑖𝑛 = 20°. For this value of ω, and the given

values of R and C, the magnitude and phase of the frequency response function are:

|𝐻(𝑗2)| =1

√1 + (𝜔𝑅𝐶)2=

1

√1 + (2 ⋅ 2Ω ⋅ 0.25𝐹)2=

1

√1 + 12=

1

√2

∠𝐻(𝑗2) = − tan−1(𝜔𝑅𝐶) = − tan−1(2 ⋅ 2Ω ⋅ 0.25𝐹) = − tan−1(1) = −45°

The output amplitude is then the product of |𝑉𝑖𝑛| and |𝐻(𝑗2)| and the output phase in the sum of ∠𝑉𝑖𝑛 and

∠𝐻(𝑗2), so that:

|𝑉𝑖𝑛| = |𝑉𝑖𝑛| ⋅ |𝐻(𝑗2)| = 3 ⋅1

√2=

1

√2

∠𝑉𝑜𝑢𝑡 = ∠𝑉𝑖𝑛 + ∠𝐻(𝑗2) = 20° + (−45°) = −25°

And the time-domain output voltage is:

𝑣𝑜𝑢𝑡(𝑡) =3

√3cos (2𝑡 − 25°)

When 𝑣𝑖𝑛 = 7 cos(4𝑡 − 60°) , 𝜔 = 4 𝑟𝑎𝑑 𝑠𝑒𝑐⁄ , |𝑉𝑖𝑛| = 7 and ∠𝑉𝑖𝑛 = −60°. For this value of ω, and the given

values of R and C, the magnitude and phase of the frequency response function are:

|𝐻(𝑗4)| =1

√1 + (𝜔𝑅𝐶)2=

1

√5

And:

∠𝐻(𝑗4) = − tan−1(𝜔𝑅𝐶) = −63.4°

The output amplitude is then the product of |𝑉𝑖𝑛| and |𝐻(𝑗4)| and the output phase is the sum of ∠𝑉𝑖𝑛 and

∠𝐻(𝑗4) so that the time-domain output voltage in this case is:

𝑉𝑜𝑢𝑡(𝑡) =7

√5cos (2𝑡 − 12.4°)



From the above examples we can see that, once the frequency response function is calculated for a circuit as a

function of frequency, we can determine the circuit’s steady-state response to any input sinusoid directly from the

frequency response function, without re-analyzing the circuit itself.

We conclude this section with one additional example, to illustrate the use of the frequency response function and

superposition to determine a circuit’s response to multiple inputs of different frequencies.

Example 10.20

Use the results of examples 10.18 and 10.19 above to determine the response vout(t) of the circuit shown below if

the input voltage is vin(t) = 3cos(2t+20) + 7cos(4t-60). Plot the input and output waveforms.

+

-0.25F

2W +

-

vin(t) vout(t)

Recall, from section 10.5, that superposition is the only valid approach for performing frequency domain analysis

of circuits with inputs at multiple frequencies. Also recall that each frequency can be analyzed separately in the

frequency domain, but that the superposition process (the summation of the individual contributions) must be

done in the time domain. For this problem, we have contributions at two different frequencies: 2 rad/sec and 4

rad/sec. Luckily, we have determined the individual responses of the circuit to these two inputs in Example 10.19.

Therefore, in the time domain, the two contributions to our output will be:

𝑣1(𝑡) =3

√2cos (2𝑡 − 25°)

And:

𝑣2(𝑡) =7

√5cos (2𝑡 − 123.4°)

The overall response is then:

𝑣𝑜𝑢𝑡(𝑡) = 𝑣1(𝑡) + 𝑣2(𝑡) =3

√2cos(2𝑡 − 25°) +

7

√5cos(2𝑡 − 123.4°)

A plot the input and output waveforms is shown below:

Time

vout(t)

vin(t)Voltage



Section Summary

• The frequency response function or frequency response describes a circuit’s input-output relationship

directly in the frequency domain, as a function of frequency. • The frequency response is a complex function of frequency H(jω) (that is, it is a complex number which

depends upon the frequency). This complex function is generally expressed as a magnitude and

phase, |𝐻(𝑗𝜔)| and ∠𝐻(𝑗𝜔) , respectively. |𝐻(𝑗𝜔)| is called the magnitude response of the circuit,

and ∠𝐻(𝑗𝜔) is called the phase response of the circuit. The overall idea is illustrated in the block diagram

below:

Input Output)j(H)j(H)j(H =

= AU f= BY

• The magnitude response of the circuit is the ratio of the output amplitude to the input amplitude. This is

also called the gain of the system. Thus, in the figure above, the output amplitude 𝐵 = |𝐻(𝑗𝜔)| ⋅ 𝐴. Note

that the magnitude response or gain of the system is a function of frequency, so that inputs of different

frequencies will have different gains.

• The phase response of the circuit is the difference between the output phase angle and the input phase

angle. Thus, in the figure above, the output phase 𝜙 = ∠𝐻(𝑗𝜔) + 𝜃. Like the gain, the phase response is a

function of frequency – inputs at different frequencies will, in general, have different phase shifts.

• Use of the frequency response to perform circuit analyses can be particularly helpful when the input

signal contains a number of sinusoidal components at different frequencies. In this case, the response of

the circuit to each individual component can be determined in the frequency domain using the frequency

response and the resulting contributions summed in the time domain to obtain the overall response.

10.5 Exercises

1. Determine the voltage across the capacitor in the circuit below if u(t) = 4cos(t+30) + 2cos(2t-45). (Hint:

this may be easier if you find the response to the input as a function of frequency, evaluate the response

for each of the above frequency components, and superimpose the results.)

u(t)+

-

2W

F4

14W

2. Determine the voltage across the resistors in the circuit below if u1(t) = 4cos(2t) and if u2(t) = cos(4t).

(Hint: this may be easier if you find the response to the input as a function of frequency, evaluate the

response for each of the above frequency components, and superimpose the results.)



u1(t)+

-

8W 8W

2H

u2(t)+

-



Real Analog Chapter 10: Lab Projects

10.4.1: Impedance

In this lab assignment, we measure impedances of resistors, capacitors, and inductors. The measured values will be

compared with our expectations based on analyses.

Before beginning this lab, you should be able to: After completing this lab, you should be able to:

Represent sinusoidal signals in phasor form Measure impedances of passive circuit elements

This lab exercise requires:

Analog Discovery module Digilent Analog Parts Kit Digital multimeter (optional)

Symbol Key:

Demonstrate circuit operation to teaching assistant; teaching assistant should initial lab notebook and grade sheet, indicating that circuit operation is acceptable.

Analysis; include principle results of analysis in laboratory report.

Numerical simulation (using PSPICE or MATLAB as indicated); include results of MATLAB numerical analysis and/or simulation in laboratory report.

Record data in your lab notebook.

General Discussion:

The concept of impedance is only appropriate in terms of the steady-state response of a circuit to a sinusoidal

input. Impedance is a complex number which provides the relationship between voltage and current phasors in

the circuit. Specifically, the impedance Z is the ratio of the voltage phasor to the current phasor:

𝑍 =𝑉

𝐼=

𝐼𝑒𝑗𝜑

𝑉𝑒𝑗𝜃 = |𝑉

𝐼| 𝑒𝑗(𝜑−𝜃) Eq. 1

where the voltage and current of interest, v(t) and i(t), are assumed to be complex exponentials of the form:

𝑣(𝑡) = 𝑉𝑒𝑗(𝜔𝑡+𝜃) = 𝑉𝑒𝑗𝜔𝑡 Eq. 2

𝑖(𝑡) = 𝐼𝑒𝑗(𝜔𝑡+𝜑) = 𝐼𝑒𝑗𝜔𝑡 Eq. 3

𝐼 and 𝑉 are phasors representing the magnitude and phase of the current and voltage, respectively. Impedance is a

very general concept which can be applied to any combination of voltage and current in a circuit. In this lab

project, however, we will be interested only in the impedance of specific circuit elements: resistors, capacitors,

and inductors.

In order to experimentally determine impedance, we must determine both voltage and current. Since oscilloscopes

do not measure current, we will use the measured voltage across a known resistance in order to infer the current

through the circuit element of interest. The appropriate circuit schematic is as shown in Fig. 1.



R

+

-

vIN(t)

i(t)

+

-

v(t)

+ -vR(t)

Circuit

element

Figure 1. Circuit used for impedance measurements.

In the circuit of Fig. 1, we can measure the voltages vR(t) and v(t). The current through the circuit element of

interest can be estimated from Ohm’s law as:

𝑖(𝑡) =𝑣𝑅(𝑡)

𝑅 Eq. 4

By measuring the voltage v(t) and estimating the current i(t) for the circuit element in Fig. 1, we can determine the

circuit element’s impedance from equation (1).

Pre-lab:

Assume that the voltages vR(t) and v(t) in Figures 2 below are of the form:

𝑣𝑅(𝑡) = 𝑣𝑅cos (𝜔𝑡 + 𝜃)

𝑣(𝑡) = 𝑉𝑐𝑜𝑠(𝜔𝑡 + 𝜑)

Determine the impedances of the impedances of the resistor R, the inductor L, and the capacitor C in

Fig. 2 below in terms of the phasor representations of the voltages vR(t) and v(t). Express your results

in terms of the magnitudes and phase angles of vR(t) and v(t).

47Ω

+ -vR(t)

+

-

v(t)

+

-

vIN(t)R

i(t)

47Ω

+ -vR(t)

+

-

v(t)

+

-

vIN(t)L

i(t)

+

-

vIN(t) C

47Ω+

-

v(t)

i(t)

+ -vR(t)

(a) (b) (c)

Figure 2. Circuits used in this lab project.

Lab Procedures:

a. Construct the circuit of Fig. 2(a) with R = 100Ω.



i. Use your function generator to apply a sinusoidal input voltage vIN(t) with an amplitude of 2V and a 0V offset. Use your oscilloscope to measure the voltages vR(t) and v(t). Set up a math channel to display the current i(t), according to equation (4). Record an image of the oscilloscope window, showing the signals vR(t), v(t), and i(t) for input signals with the following frequencies:

1kHz

5kHz

10kHz

ii. For each of the above three frequencies, tabulate: the amplitudes of v(t) and i(t), and the time difference between v(t) and i(t).

iii. Calculate the impedance of the resistor at the above three frequencies. Compare your results to your expectations from the pre-lab analyses. Include a percent difference between your expectations and your measured impedances. Note: Appendix A of this lab assignment provides tips relative to gain and phase measurement of sinusoidal signals.

iv. Demonstrate operation of your circuit to the TA and have them initial the appropriate pages of your lab notebook and the lab worksheet.

b. Construct the circuit of Fig. 2(b) with L = 1mH. i. Use your function generator to apply a sinusoidal input voltage vIN(t) with an amplitude of 2V

and a 0V offset. Use your oscilloscope to measure the voltages vR(t) and v(t). Set up a math channel to display the current i(t), according to equation (4). Record an image of the oscilloscope window, showing the signals vR(t), v(t), and i(t) for input signals with the following frequencies:

1kHz

5kHz

10kHz


iii. Calculate the impedance of the inductor at the above three frequencies. Compare your results to your expectations from the pre-lab analyses. Include a percent difference between your expectations and your measured impedances. Note: Appendix A of this lab assignment provides tips relative to gain and phase measurement of sinusoidal signals.


c. Construct the circuit of Figure 2(c) with C = 100nF. i. Use your function generator to apply a sinusoidal input voltage vIN(t) with an amplitude of 2V

and a 0V offset. Use your oscilloscope to measure the voltages vR(t) and v(t). Set up a math channel to display the current i(t), according to equation (4). Record an image of the oscilloscope window, showing the signals vR(t), v(t), and i(t) for input signals with the following frequencies:

1kHz

5kHz

10kHz


iii. Calculate the impedance of the capacitor at the above three frequencies. Compare your results to your expectations from the pre-lab analyses. Include a percent difference between



your expectations and your measured impedances. Note: Appendix A of this lab assignment provides tips relative to gain and phase measurement of sinusoidal signals.




Appendix A: Measuring Gain and Phase:

The gain of a system at a particular frequency is the ratio of the magnitude of the output voltage to the magnitude

of the input voltage at that frequency, so that:

𝐺𝑎𝑖𝑛 =∆𝑉𝑜𝑢𝑡

∆𝑉𝑖𝑛

Where ∆𝑉𝑜𝑢𝑡 and ∆𝑉𝑖𝑛 can be measured from the sinusoidal input and output voltages as shown in the figure

below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

DVin

DVout

The phase of a system at a particular frequency is a measure of the time shift between the output and input

voltage at that frequency, so that:

𝑃ℎ𝑎𝑠𝑒 =∆𝑇

𝑇× 360°

where ∆𝑇 and ∆𝑇 can be measured from the sinusoidal input and output voltages as shown in the figure below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

T

DT



Real Analog Chapter 10: Lab Worksheets

10.4.1: Impedance (75 points total)

a. Resistor (25 points)

1. In the space below, provide the impedances for the resistor in Fig. 2(a), in terms of the magnitudes and phase angles of vR(t) and v(t). (4 pts)

2. Attach to this worksheet images of the oscilloscope main window, showing the signals vR(t), v(t), and i(t) of Fig. 2(a) for input signals with the following frequencies:1kHz, 5kHz, and 10kHz. (9 pts, 3pts per image)

3. In the space below, provide a table which gives the amplitude difference and time shift between v(t) and i(t) for the circuit of Fig. 2(a) at frequencies of 1kHz, 5kHz, and 10kHz. (3 pts)

4. In the space below, provide the measured impedance of the resistor at the three frequencies of interest. Compare your results to your expectations from the pre-lab, including percent differences between measured and expected impedances. (6 pts)

5. DEMO: Have a teaching assistant initial this sheet, indicating that they have observed your system’s operation. (3 pts total)

TA Initials: _______



b. Inductor (25 points)

1. In the space below, provide the impedances for the inductor in Fig. 2(b), in terms of the magnitudes and phase angles of vR(t) and v(t). (4 pts)

2. Attach to this worksheet images of the oscilloscope main window, showing the signals vR(t), v(t), and i(t) of Fig. 2(b) for input signals with the following frequencies:1kHz, 5kHz, and 10kHz. (9 pts, 3pts per image)

3. In the space below, provide a table which gives the amplitude difference and time shift between v(t) and i(t) for the circuit of Fig. 2(b) at frequencies of 1kHz, 5kHz, and 10kHz. (3 pts)

4. In the space below, provide the measured impedance of the inductor at the three frequencies of interest. Compare your results to your expectations from the pre-lab, including percent differences between measured and expected impedances. (6 pts)





c. Capacitor (25 points)

1. In the space below, provide the impedances for the capacitor in Fig. 2(c), in terms of the magnitudes and phase angles of vR(t) and v(t). (4 pts)

2. Attach to this worksheet images of the oscilloscope main window, showing the signals vR(t), v(t), and i(t) of Fig. 2(c) for input signals with the following frequencies:1kHz, 5kHz, and 10kHz. (9 pts, 3pts per image)

3. In the space below, provide a table which gives the amplitude difference and time shift between v(t) and i(t) for the circuit of Fig. 2(c) at frequencies of 1kHz, 5kHz, and 10kHz. (3 pts)

4. In the space below, provide the measured impedance of the capacitor at the three frequencies of interest. Compare your results to your expectations from the pre-lab, including percent differences between measured and expected impedances. (6 pts)






10.6.1: Passive RL Circuit Response

In this lab assignment, we will be concerned with the steady-state response of electrical circuits to sinusoidal

inputs. Figure 1(a) shows a block-diagram representation of the system. The input and output signals both have

the same frequency, but the two signals can have different amplitudes and phase angles.

The analysis of the circuit of Fig. 1(a) can be simplified by representing the sinusoidal signals as phasors. The

phasors provide the amplitude and phase information of sinusoidal signals. By comparing the phasors representing

the input and output signals, the effect of the circuit can be represented as an amplitude gain between the output

and input signals and a phase difference between the output and input signals, as shown in Fig. 1(b).

CircuitInput

u(t)=Acos(t+)

Output

y(t)=Bcos(t+f)

(a) Physical circuit

Input

U=A

Output

Y=Bf( )f −

A

B

(b) Phasor representation of circuit input-output relationship.

Figure 1. Steady-state sinusoidal circuit analysis

In this lab assignment, we will measure the gain and phase responses of a passive RL circuit and compare these

measurements with expectations based on analysis.


Represent sinusoidal signals in phasor form Determine electrical circuit steady-state

sinusoidal responses in phasor form

Measure phasor form of circuit steady-state sinusoidal response

Measure input impedance of electrical circuit



Symbol Key:







General Discussion:

Consider the RL circuit shown in Fig. 2 below. The input to the circuit is an applied voltage and we choose the

current supplied by the source to be the system output. The differential equation relating the applied voltage vIN(t)

to the input current iIN(t) can be obtained by applying KVL around the single loop:

If we assume that the input voltage and current are complex exponentials of the form:

𝑣𝐼𝑁(𝑡) = 𝑉𝑒𝑗(𝜔𝑡+𝜃) Eq. 1

𝑖𝐼𝑁(𝑡) = 𝐼𝑒𝑗(𝜔𝑡+𝜑) Eq. 2

We can write the circuit’s input-output relation as a ratio between the current and the voltage:

𝐼

𝑉=

𝐼𝑒𝑗𝜑

𝑉𝑒𝑗𝜃 =1

𝑅+𝑗𝜔𝐿 Eq. 3

Where 𝐼 and 𝑉 are phasors representing the magnitude and phase of the input current and input voltage to the

circuit, respectively. This input-output relation can be written in terms of an amplitude gain and a phase shift:

|𝐼

𝑉| = |

1

𝑅+𝑗𝜔𝐿| =

1

√𝑅2+𝜔2𝐿2 Eq. 4

𝜑 − 𝜃 = − tan−1 (𝜔𝐿

𝑅) Eq. 5

R

+ -vR(t)

+

-

vL(t)

+

-

vIN(t)

iIN(t)

L

Figure 2. RL circuit.

Pre-lab:

a. Show that the amplitude gain and phase difference between the input voltage and the input current are as shown in equations (4) and (5).

b. The cutoff frequency for the circuit of Fig. 2 is given to be 𝜔𝑐 =𝑅

𝐿. Calculate the cutoff frequency

for the circuit of Fig. 2 if L = 1mH and R = 47Ω.

c. Determine the gain and phase difference for the RL circuit for frequencies 𝜔 ≈ 0, 𝜔 → ∞, and

𝜔 = 𝜔𝑐 if L = 1mH and R = 47Ω. d. Do your low and high frequency gain results in part (c) agree with your expectations based on the

inductor’s low and high frequency behavior? (e.g. calculate the inductor impedance at low and high frequencies, substitute these impedances into the circuit of Fig. 2, calculate the response of the resulting resistive circuit, and compare to the results of part (c).)

Notes:

In this lab assignment, we will measure vIN(t) and vL(t). These measurements will be used to estimate the gain and

phase difference between vIN(t) and iIN(t) and the gain and phase difference between vL(t) and iIN(t). These results

will be compared with our expectations based on the pre-lab analyses. We do not have the ability to directly

measure a time-varying current, so we will infer iIN(t) by measuring vIN(t) - vL(t) and determining iIN(t) by:



𝑖𝐼𝑁(𝑡) =𝑣𝐼𝑁(𝑡)−𝑣𝐿(𝑡)

𝑅 Eq. 6

All signals we will be dealing with are sinusoidal. Appendix A of this lab assignment provides tips relative to gain

and phase measurement of sinusoidal signals.

Lab Procedures:

Construct the circuit of Fig. 2 with L = 1mH and R = 47Ω.

a. Use your function generator to apply a sinusoidal input at vIN(t). Use your oscilloscope to display both vIN(t) and vL(t). Use the oscilloscope’s math operation to display the input current, iIN(t), as provided by equation (6). Record the amplitude of vIN(t) and iIN(t) and the time delay between vIN(t) and iIN(t) for the following input voltage frequencies:

• 𝜔 ≈𝜔𝑐

10 (low frequency input)

• 𝜔 ≈ 10𝜔𝑐 (high frequency input)

• 𝜔 ≈ 𝜔𝑐 (corner frequency input)

b. Demonstrate operation of your circuit to the TA and have them initial the appropriate page(s) of your lab notebook and the lab worksheet.

c. Calculate the measured gains and phase differences between iIN(t) and vIN(t) for the three frequencies listed in part (b) above. Compare your measured results with your expectations from the pre-lab. Comment on your results.

Appendix A: Measuring Gain and Phase



Gain =∆𝑉𝑜𝑢𝑡

∆𝑉𝑖𝑛

where ∆𝑉𝑜𝑢𝑡 and ∆𝑉𝑖𝑛 can be measured from the sinusoidal input and output voltages as shown in the figure

below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

DVin

DVout





Phase=∆𝑇

𝑇× 360°

where ∆𝑉𝑖𝑛 and T can be measured from the sinusoidal input and output voltages as shown in the figure below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

T

DT




10.6.1: Passive RL Circuit Response (40 points total)

1. Attach, to this worksheet, your derivation of gain and phase expressions (equations (5) and (6)) (3 pts)

2. In the space below, provide the cutoff frequency calculated in part (b) of the pre-lab. (3 pts)

3. In the space below, provide the gain (𝐼

𝑉) and phase (∠𝐼 − ∠𝑉) for the RL circuit at low, high, and corner

frequencies as determined from part (c) of your pre-lab analysis. (9 pts)

4. Comment below on the inductor physical behavior at low and high frequencies vs. expressions provided in (2) above. (2 pts)

5. In the space below, provide a table listing vIN(t) and iIN(t) and time delays between vIN(t) and iIN(t) for the three frequencies of interest in part (a) of the lab procedures. (10 pts)



6. In the space below, provide a table listing the measured gains and phase differences between iIN(t) and vIN(t) and iIN(t) and vL(t) for the three frequencies of interest. (8 pts)






10.6.2: Passive RC Circuit Response








CircuitInput

u(t)=Acos(ω t + θ)

Output

y(t)=Bcos(ω t + ϕ)

Input

U=A

Output

Y=Bf( )f −

A

B

In this lab assignment, we will measure the gain and phase responses of a passive RC circuit and compare these

measurements with expectations based on analysis. These measurements will be used to estimate the impedance

of the overall RC circuit.


Represent sinusoidal signals in phasor form Determine electrical circuit steady-state

sinusoidal responses in phasor form







Symbol Key:





General Discussion:

In this lab assignment, we will determine the input impedance of the passive RC circuit shown in Fig. 1. The input

impedance of a circuit is defined as the ratio of input voltage to input current. Thus, for the circuit of Fig. 1, the

input impedance is represented in phasor form as:

𝑍𝐼𝑁 =𝑉𝐼𝑁

𝐼𝐼𝑁 Eq. 1

Where 𝑉𝐼𝑁 is the phasor representation of the circuit input voltage and 𝐼𝐼𝑁 is the phasor representation of the

input current to the circuit.

The cutoff frequency for the circuit of Fig. 1 is:

𝜔𝑐 =1

𝑅𝐶 Eq. 2

R

+

-

vIN(t)

iIN(t)

C

Figure 1. Passive RC circuit.

Pre-lab:

a. Determine an expression for the input impedance of the circuit of Fig. 1 in terms of R, C, and ω.

b. If R = 100W and C = 1mF, determine the cutoff frequency for the circuit. Also determine the input impedance for frequencies of:

• 𝜔 =𝜔𝑐


• 𝜔 = 10𝜔𝑐 (high frequency input)

• 𝜔 = 𝜔𝑐 (corner frequency input)

c. Check your low and high frequency results in part (b) relative to your expectations based on the capacitor’s low and high frequency behavior.

Lab Procedures:

Construct the circuit of Fig. 3, using R = 100W and C = 1mF.



a. Measure the input voltage amplitude, the input current amplitude, and the time delay between the input voltage and the input current for the following frequencies:

• 𝜔 ≈𝜔𝑐


• 𝜔 ≈ 10𝜔𝑐 (high frequency input)

• 𝜔 ≈ 𝜔𝑐 (corner frequency input)

Use your data to calculate the input impedance (magnitude and phase) of the circuit for the

above frequencies. Create a table providing the measured data and the calculated input

impedances at the above frequencies.

b. Compare your measured results with your expectations based on the analysis you did in the pre-lab.

c. Demonstrate operation of your circuit to the TA and have them initial the appropriate page(s) of your lab notebook and the lab worksheet.

Hint:

The process to perform the above lab procedures is comparable to the process performed in lab assignment

10.6.1. Be sure to record all necessary data and any calculations you perform to obtain your results in your lab

notebook. Appendix A of this lab assignment provides tips relative to gain and phase measurement of sinusoidal

signals.





∆𝑉𝑖𝑛

where ∆𝑉𝑜𝑢𝑡 and ∆𝑉𝑖𝑛 can be measured from the sinusoidal input and output voltages as shown in the figure

below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

DVin

DVout





Phase =∆𝑇

𝑇× 360°

where ∆𝑇 and T can be measured from the sinusoidal input and output voltages as shown in the figure below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

T

DT




10.6.2: Passive RC Circuit Response (30 points total)

1. In the space below, provide your expression for the input impedance of the circuit. (3 pts)

2. In the space below, provide the cutoff frequency calculated in part (b) of the pre-lab and the circuit’s input impedance at the three specified frequencies. (3 pts)

3. In the space below, comment on your results in part 2 above, relative to your expectations based on the capacitor’s behavior at low and high frequencies. (2 pts)

4. In the space below, provide a table listing the measured input and output voltage amplitudes, the time difference between the input voltage and input current, and the calculated input impedances at the three frequencies of interest. (10 pts)



5. In the space below, provide a brief comparison between your measured results and your expectations from the pre-lab. Include a percent difference between the expected and measured impedances at the three frequencies of interest. (7 pts)






10.6.3: Inverting Voltage Amplifier








CircuitInput

u(t)=Acos(t+)

Output

y(t)=Bcos(t+f)


Input

U=A

Output

Y=Bf( )f −

A

B



In this lab assignment, we will measure the gain and phase responses of an inverting voltage amplifier circuit and

compare these measurements with expectations based on analysis.


Represent sinusoidal signals in phasor form Represent electrical circuit steady-state

sinusoidal responses in phasor form Analyze operational amplifier-based circuits





Symbol Key:







General Discussion:

In this lab assignment, we will measure the frequency domain input-output relation governing the inverting

voltage amplifier shown in Fig. 1. The frequency domain input-output relation for the circuit of Fig. 1 is:

𝑉𝑂𝑈𝑇

𝑉𝐼𝑁= −

1

𝑗𝜔𝑅𝐶+1 Eq. 1

So that the amplitude gain between the output and input is:

|𝑉𝑂𝑈𝑇

𝑉𝐼𝑁| = −

1

√(𝜔𝑅𝐶)2+1 Eq. 2

And the phase difference between the output and input is:

∠𝑉𝑂𝑈𝑇 − ∠𝑉𝐼𝑁 = 180° − tan−1 (1

𝜔𝑅𝐶) Eq. 3

+

-

R

R

VIN(t)+

-

VOUT(t)

C

+

-

Figure 1. Inverting voltage amplifier.

Pre-lab:

a. Show that equation (1) is the input-output relation for the circuit of Fig. 1. Also verify equations (2) and (3) above.

b. Determine the cutoff frequency of the circuit if R = 10kW and C = 10nF. Also determine the amplitude gain and the phase difference between the circuit’s input and output voltages for the circuit1. Also determine the input impedance for frequencies of:

• 𝜔 =𝜔𝑐


• 𝜔 = 10𝜔𝑐 (high frequency input)

• 𝜔 = 𝜔𝑐 (corner frequency input)


Lab Procedures:

Construct the circuit of Fig. 2, using R = 10kW and C = 10nF.

a. Use the waveform generator to apply a sinusoidal signal with 2V amplitude and 0V offset to the circuit. Set up the oscilloscope to measure both the input and output voltages. Measure the

1 Be sure to use units of radians/second for in equations (2) and (3)!



amplitudes of the input and output voltage signal, and the time delay between the input and output signal for inputs with the following frequencies:

100 Hz 1 KHz 5 KHz

b. Record an image of the oscilloscope window, showing the signals VIN(t) and VOUT(t), for each of the above frequencies.

c. Use your measurements to calculate the amplitude gain and phase difference of the circuit for the above three frequencies. Compare your measured results with your expectations based on the analysis you did in the pre-lab.

d. Demonstrate operation of your circuit to the TA and have them initial the appropriate page(s) of your lab notebook and the lab worksheet.

Hint:

Be sure to record all necessary data and any calculations you perform to obtain your results in your lab notebook.

Appendix A of this lab assignment provides tips relative to gain and phase measurement of sinusoidal signals.





∆𝑉𝑖𝑛


below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

DVin

DVout



Phase =∆𝑇

𝑇× 360°

Where ∆𝑇 and T can be measured from the sinusoidal input and output voltages as shown in the figure below.



Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

T

DT




10.6.3: Inverting Voltage Amplifier (45 points total)

1. Attach, to this worksheet, your derivation of equations (1), (2), and (3). (3 pts)

2. In the space below, provide the cutoff frequency calculated in part (b) of the pre-lab. (3 pts)

3. In the space below, provide the gain, phase, and impedance for the circuit at low, high, and corner frequencies as determined from part (c) of your pre-lab analysis. (6 pts)

4. Comment below on the capacitor physical behavior at low and high frequencies vs. expressions provided in (3) above. (2 pts)

5. Attach to this worksheet images of the oscilloscope window, showing the input and output voltages as functions of time for each of the three specified frequencies. (6 pts, 2 pts each image)



6. In the space below, tabulate the amplitudes of the input and output voltages, the time difference between the input and output voltages, the gain and phase of the circuit, and the circuit’s input impedance, for each of the three frequencies of interest in part (a) of the lab procedures. (12 pts)

7. In the space below, comment on the differences between the measured and expected gain and phase of the circuit at each of the frequencies in part 6 above. (e.g. compare your expressions in part 3 above with the measured data). (8 pts)






10.6.4: Non-inverting Voltage Amplifier








CircuitInput

u(t)=Acos(t+)

Output

y(t)=Bcos(t+f)


Input

U=A

Output

Y=Bf( )f −

A

B



In this lab assignment, we will measure the gain and phase responses of a non-inverting voltage amplifier circuit

and compare these measurements with expectations based on analysis.


Represent sinusoidal signals in phasor form Represent electrical circuit steady-state

sinusoidal responses in phasor form Analyze operational amplifier-based circuits





Symbol Key:







General Discussion:

In this lab assignment, we will measure the frequency domain input-output relation governing the voltage

amplifier shown in Fig. 1. The frequency domain input-output relation for the circuit of Fig. 1 is:

𝑉𝑂𝑈𝑇

𝑉𝐼𝑁=

𝑅1+𝑅2

𝑅1

1

𝑅3𝐶

𝑗𝜔+1

𝑅3𝐶

Eq. 1

So that the amplitude gain and phase difference between the output and input are:

|𝑉𝑂𝑈𝑇

𝑉𝐼𝑁| = 2

1

𝑅𝐶

√𝜔2+(1

𝑅𝐶)

2 Eq. 2

∠𝑉𝑂𝑈𝑇 − ∠𝑉𝐼𝑁 = − tan−1(𝜔𝑅𝐶) Eq. 3

-

+

R

R

+

-C

+

-

RVin

Vout

Figure 1. Non-inverting voltage amplifier.

Pre-lab:

a. Show that equation (1) is the input-output relation for the circuit of Fig. 1. Also verify equations (2) and (3) above.

b. If R = 10kW and C = 10nF, determine the amplitude gain and the phase difference between the circuit’s input and output voltages for the circuit for input frequencies of 100Hz, 5kHz, and 10kHz2.


Lab Procedures:

Construct the circuit of Fig. 2, using R = 10kW and C = 10nF.

a. Use the waveform generator to apply a sinusoidal signal with 1V amplitude and 0V offset to the circuit. Set up the oscilloscope to measure both the input and output voltages. Measure the amplitudes of the input and output voltage signal, and the time delay between the input and output signal for inputs with the following frequencies: 100 Hz 5 KHz 10 KHz

2 Be sure to use units of radians/second for when evaluating equations (2) and (3)!



b. Record an image of the oscilloscope window, showing the signals VIN(t) and VOUT(t), for each of the above frequencies.

c. Use your measurements to calculate the amplitude gain and phase difference of the circuit for the above three frequencies. Compare your measured results with your expectations based on the analysis you did in the pre-lab.

d. Demonstrate operation of your circuit to the TA and have them initial the appropriate page(s) of your lab notebook and the lab worksheet.

Hint:

Be sure to record all necessary data and any calculations you perform to obtain your results in your lab notebook.

Appendix A of this lab assignment provides tips relative to gain and phase measurement of sinusoidal signals.

Appendix A: Measuring Gain and Phase




∆𝑉𝑖𝑛


below.

Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

DVin

DVout



Phase =∆𝑇

𝑇× 360°

Where ∆𝑇 and T can be measured from the sinusoidal input and output voltages as shown in the figure below.



Time

Voltage

Input

voltage, Vin

Output

voltage, Vout

T

DT




10.6.4: Non-inverting Voltage Amplifier (40 points total)

1. Attach, to this worksheet, your derivation of equations (1), (2), and (3). (3 pts)

2. In the space below, provide the calculated gain and phase for the circuit at frequencies of 100Hz, 5kHz, and 10kHz. (Part (b) of the pre-lab.) (4 pts)

3. Comment below on the capacitor physical behavior at low and high frequencies vs. expressions provided in (2) above. (2 pts)

4. Attach to this worksheet images of the oscilloscope window, showing the input and output voltages as functions of time for each of the three specified frequencies. (6 pts, 2 pts each image)



5. In the space below, tabulate the amplitudes of the input and output voltages, the time difference between the input and output voltages, and the gain and phase of the circuit for each of the three frequencies of interest in part (a) of the lab procedures. (12 pts)

6. In the space below, comment on the differences between the measured and expected gain and phase of the circuit at each of the frequencies in part 6 above. (e.g. compare your expressions in part 3 above with the measured data). (8 pts)





Real Analog Chapter 10: Homework

10.1 A circuit is described by the differential equation:

2𝑑𝑖(𝑡)

𝑑𝑡+ 10𝑖(𝑡) = 10𝑣(𝑡)

If v(t) = 3cos(5t), determine the steady-state response of i(t).

10.2 The differential equation governing a circuit is:

3𝑑𝑖(𝑡)

𝑑𝑡+ 6𝑖(𝑡) = 𝑣𝑠(𝑡)

Where vs(t) is the input and i(t) is the output. Determine the steady-state response of the circuit to an input

𝑣𝑠(𝑡) = 5cos (4𝑡 + 30°).

10.3 For the circuit below,

a. The equivalent impedance seen by the source.

b. iC(t), t→.

iC(t)

4cos(3t+30) +-

H3

1

4Ω

F6

1

10.4 For the circuit below, find v(t), t→.

10cos(4t) +-

H2

1 4Ω

F8

1

H2

1

4Ωv(t)

+

-

10.5 For the circuit shown, find


b. The steady-state response of the voltage across the resistor, vR(t).



20cos(10t)

0.1H

2ΩF30

1vR(t)

+

-



b. The steady-state current delivered by the source.

+

-2cos(3t) H

3

1

F9

1

3Ω

1H

10.7 For the circuit shown, determine


b. The steady-state current out of the source, is(t→).

+

-3cos(2t)V

is(t) 6Ω

1H

3Ω

F6

1



b. iS(t), t→.

+

-5cos(4t+30)

2Ω

4Ω

F16

1

1H

is(t)





b. vR(t→).

+

-2cos(4t+25)

4Ω

2Ω

0.5H

F8

1+ -vR(t)

10.10 For the circuit shown, find is(t), t>0.

+

-5cos(2t-30)V

is(t)

2Ω

1H

0.5F 2Ω



b. The steady-state response of the voltage across the resistor, vR(t).

20cos(10t)

0.1H

2ΩF30

1vR(t)

+

-

Real Analog Chapter 10: Steady-state Sinusoidal Analysis ...

Documents